tl;dr- Leading zeros aren't significant because they trivially drop out. Still, if you prefer to keep 'em, that's fine, too; you'll just end up having a bunch of leading zeros.
And, no, this isn't something anyone should really want to do. But, the logic behind it and the consequences probably help show why.
Background: Regarding the construction of numbers
First, to define numbers:
Natural numbers are defined through enumeration on a number line from $0 .$
Integers as defined as natural numbers extended with decrementation (inverse enumeration) on a number line, allowing negative values.
Real numbers are defined as integers with interpolation, allowing decimal values.
Conceptually, it'd be simplest if we gave each integer its own, unique symbol. But since no one wants to have to memorize arbitrarily many symbols, we tend to construct numeric identifiers through a transform$$
n
~~ \Rightarrow ~~
\sum_{i} c_i \cdot {b}^{i}
\, ,$$where
$c_i$ is a numeric symbol selected from a limited subset of "digits" $\in \left[0,~b\right) ;$
$b$ is the base (and usually $10$);
then emit this construction as a string,$$
\hspace{25px}
\boxed{\begin{alignat}{7}
&\texttt{for}~\left(\texttt{var}~i~=~i_{\text{max}};~~~i~\ge~i_{\text{min}};~~~i\text{--}\right) \\
&\{ \\
& \hspace{2em} \texttt{Print} \left( c_i \right) ; \\
\\
& \hspace{2em} \texttt{if} \left( i ~\text{==}~ 0 \right) \\
& \hspace{2em} \{ \\
& \hspace{4em} \texttt{Print} \left( `` . " \right) ; \\
& \hspace{2em} \} \\
& \}
\end{alignat}}
_{~ \large{.}}$$
Background: Regarding multiplication
Since multiplication is distributive, then the product of two numbers written in the same base, $b ,$ is$$
\begin{alignat}{7}
n^{\text{A}} \times n^{\text{B}} ~~ & \Rightarrow ~~ &&
\left( \sum_{i} c_i^{\text{A}} \cdot {b}^{i} \right) \times \left( \sum_{j} c_j^{\text{B}} \cdot {b}^{j} \right) \\[5px]
& = && \sum_{i} {\sum_{j} {c_i^{\text{A}} \cdot c_j^{\text{B}} \cdot {b}^{i} \cdot {b}^{j}}} \\[5px]
& = && \sum_{i} {\sum_{j} {c_i^{\text{A}} \cdot c_j^{\text{B}} \cdot {b}^{i+j}}}
\, .\end{alignat}
$$
Background: Regarding truncation
The above definitions are written for numbers that contain infinite information. In practice, computers (including humans) are finite (unless you find a hypercomputer), so we terminate the procedure at two ends:
We declare some minimal basis, $\cdot b^{i_{\text{min}}} ,$ past which we ignore all further bases, typically under the argument that they're noisy (if from measurement/estimation) or to save on computation work (as computers do).
We declare some maximal basis, $\cdot b^{i_{\text{max}}} ,$ past which we ignore all further bases. Usually, we choose to selected $i_{\text{max}}$ such that we truncate only terms in which $c_i = 0 ,$ since the zero-terms don't affect anything, anyway.
First, we note that any basis $\cdot b^k$ is affected by a noisy term if $k < i_{\text{min}} + j_{\text{max}}$ or/and $k < i_{\text{max}} + j_{\text{min}}$ – ignoring cases in which $c_i c_j \geq b ,$ which I'll mention later.
Second, we note that any basis $\cdot b^k$ is a zero-term if $k > i_{\text{max}} + j_{\text{max}} .$
Given these two constraints, we're therefore only interested in bases$$
\cdot b^k ~~ \text{such that} ~~ k \in \left[ \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} , ~ i_{\text{max}} + j_{\text{max}} \right]
\, .$$Since the number of elements in an inclusive range like this, i.e. $\left[ n_{\text{min}}, ~ n_{\text{max}}\right] ,$ is $1 + n_{\text{max}} - n_{\text{min}} ,$we're therefore interested in$$
\begin{alignat}{7}
1 + i_{\text{max}} + j_{\text{max}} - \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)}
~~ & = ~~ && 1 + \min{\left( i_{\text{max}} - i_{\text{min}} , ~ j_{\text{max}} - j_{\text{min}} \right)} \\[5px]
& = && \min{\left(1+ i_{\text{max}} - i_{\text{min}} , ~1+ j_{\text{max}} - j_{\text{min}} \right)}
\end{alignat}
$$bases.
So, ya know how they say that, when you multiply two numbers with significant figures, the product has the lesser of the multiplicands' significant figures? That's because$$
\underbrace{1 + k_{\text{max}} - k_{\text{min}}}_{\begin{array}{c} \text{significant figures} \\ \text{in the product} \end{array}}
~~ = ~~ \min{(\underbrace{1+ i_{\text{max}} - i_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{first multiplicand} \end{array}} , ~ \underbrace{1+ j_{\text{max}} - j_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{second multiplicand} \end{array}} )}
\, .$$This is, the product has the lesser of the number of significant figures from either multiplicand.
Except, there's one problem here: the above logic assumed that bases don't overflow. Which would be true if we were working in Base-2 (binary), but in Base-10 (decimal), we have cases in which $c_i c_j \geq b ,$ e.g. $5 \times 5 \geq 10 .$ Won't discuss that here since it's ignored by the standard rules, but I think the problem's obvious enough. That said, significant figures are meant to be an easy trick rather than used for rigorous calculations, so that they're a bit broken is kind of a given. (See also: my answer here.)
Considering leading-zeros significant
In the above derivation of significant-figure logic, we selected the rule that leading zeros are to be ignored. So, what happens if we do consider them to be significant?
Specifically, you're asking about the case in which $i_{\text{max}}$ or $j_{\text{max}}$ is less-than-zero – e.g., as in $0.01 ,$ in which $i_{\text{max}} = -2$ – and then asking why we can't consider the zeros that we still wrote, so presumably $i_{\text{max}} = 0 .$
So, let's call your alterations $i_{\text{max}}^{*}$ and $j_{\text{max}}^{*} ,$ where$$
i_{\text{max}}^{*} ~ \equiv ~\max{\left(i_\text{max}, ~ 0\right)}
~~~~ \text{and} ~~~~
j_{\text{max}}^{*} ~ \equiv ~\max{\left(j_\text{max}, ~ 0\right)}
\, .$$Then, we say that we're interested in "significant" figures that include leading zeros, though we must still reference the prior notions of $i_{\text{max}}$ and $j_{\text{max}}$ because they're important to the issue of tracking the propagation of noise in the calculation.
So then, we're interested in the bases$$
\cdot b^k ~~ \text{such that} ~~ k \in \left[ \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} , ~ i_{\text{max}}^{*} + j_{\text{max}}^{*} \right]
\, ,$$which, to redo the element-count calculation, contains$$
1
+ i_{\text{max}}^{*} + j_{\text{max}}^{*}
- \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)}
$$members.
So, to derive the new rule for significant figures, we just need to rewrite $i_{\text{min}}$ and $j_{\text{max}}$ in terms of $i_{\text{min}}^{*}$ and $j_{\text{max}}^{*} ,$ and we're done.
So, um. If $i_{\text{max}}^{*} ~ \equiv ~\max{\left(i_\text{max}, ~ 0\right)} ,$ and we know $i_{\text{max}}^{*} ,$ then how do we calculate $i_\text{max} ?$
I mean, obviously, $i_{\text{max}}$ is either going to be equal to $i_{\text{max}}^{*}$ if $i_{\text{max}}^{*} > 0 ,$ but if $i_{\text{max}}^{*} = 0 ,$ then all we know is that $i_{\text{min}} \leq i_{\text{max}} \leq 0 .$ Our problem is that this information is lost, such that merely knowing the number of "significant figures" is insufficient to establish how many we need.
But, screw it, significant figures are a hack anyway. And, so long as we respect the bound on the least-significant basis, we can keep extra zeros if we like. Since, ya know, they don't affect anything.
So to avoid accidentally truncating leading non-zero digits, it's left to us to write the rules such that there're at least as many significant figures in the product as necessary to keep it consistent.
So, we need a number of significant figures equal to$$
\max{\left(
1
+ i_{\text{max}}^{*} + j_{\text{max}}^{*}
- \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)}
~~~~\forall
\begin{array}{l}
i_{\text{max}} \in \left[i_{\text{min}}, ~ \max{\left(0, ~ i_{\text{max}}^* \right)}\right] \\
j_{\text{max}} \in \left[j_{\text{min}}, ~ \max{\left(0, ~ j_{\text{max}}^* \right)}\right]
\end{array}
\right)}
\, ,$$which reduces to$$
1 + i_{\text{max}}^{*} - i_{\text{min}} + j_{\text{max}}^{*} - j_{\text{min}}
\, ,$$or$$
\underbrace{1 + i_{\text{max}}^{*} - i_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{first multiplicand} \end{array}}
+ \underbrace{1 + j_{\text{max}}^{*} - j_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{second multiplicand} \end{array}}
- 1
\, .$$In other words, the new rule is that we need to retain a number of significant figures equal to the sum of the significant figures of the multiplicands, minus one, where the new "significant" digits, if any, are leading zeros.
Which is a rule you can have, but it then requires that you recount the significant figures of the product afterward before doing further operations, as this logic's non-conservative.
Problem: Number of significant figures grows
To avoid improper truncation, we had to keep at least as many leading zeros as to ensure that nothing was dropped. If these calculations are repeated, the bloat of leading zeros may continue to grow. (Which I haven't actually worked out; typing this all out took longer than I originally estimated, and, honestly, I'm bored. =P)
But, since the premise of this derivation is that we don't mind doing trivial work, being why we rejected dropping terms $c_i c_j b^{i+j} = 0$ from the calculation, that's presumably not an issue for someone who'd want to use this logic.
Conclusion
Long story short, you can consider leading zeros "significant" if you want, then maintain a bunch of leading zeros in front of numbers to maintain that logic.
It doesn't really mean anything, as it's basically just keeping extra zeros, but it's a mathematically consistent calculation approach one could take if they were so inclined.
