Explicit indices
The difficulty here arises from making indices implicit in tensor product expressions. For example, the unitary $U$ corresponding to controlled-NOT gate on qubits $1$ and $2$ and identity on qubit $3$ is often written down as
$$
U=\text{CNOT}\otimes I\tag1
$$
but a similar unitary $U'$ corresponding to controlled-NOT on qubits $1$ and $3$ and identity on qubit $2$ cannot be written this way
$$
U'=\text{???}.\tag2
$$
A solution is to introduce subscripts that identify the qubits
$$
U_{123}=\text{CNOT}_{12}\otimes I_3\tag{1'}
$$
which allows us to write $(2)$ as
$$
U_{123}'=\text{CNOT}_{13}\otimes I_2.\tag{2'}
$$
Computing matrix elements of tensor products
The additional advantage of this more explicit notation is that it provides a simple algorithm for computing matrix elements of tensor products. To that end replace qubit list like $123$ with two lists of indices like $i_1i_2i_3;k_1k_2k_3$ where $i_m$ are row indices and $k_m$ are column indices and where $m\in\{1,2,3\}$ identifies a qubit corresponding to each index. Then $(2)$ becomes
$$
U_{i_1i_2i_3;k_1k_2k_3}=\text{CNOT}_{i_1i_2;k_1k_2}I_{i_3;k_3}\tag3
$$
and $(2')$ becomes
$$
U_{i_1i_2i_3;k_1k_2k_3}'=\text{CNOT}_{i_1i_3;k_1k_3}I_{i_2;k_2}.\tag{3'}
$$
If we now interpret $i_1i_2i_3\in\{000,001,\dots,111\}$ as the binary encoding of the row number of the matrix on the LHS and $k_1k_2k_3$ as the binary encoding of the column number, then $(3)$ and $(3')$ give us simple direct formulas for computing matrix elements of $U$ and $U'$ from the matrix elements of the factor matrices $\text{CNOT}$ and $I$.
Example
For example, the matrix of the CNOT gate is
$$
\text{CNOT}=\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&0&1\\
0&0&1&0
\end{bmatrix}.\tag4
$$
In other words,
$$
\text{CNOT}_{ij;i'j'}=\begin{cases}
1&\text{ if }i=i'=0\text{ and }j=j',\\
1&\text{ if }i=i'=1\text{ and }j\ne j',\\
0&\text{ otherwise}.
\end{cases}\tag5
$$
Also,
$$
I_{i;j}=\begin{cases}
1\text{ if }i=j,\\
0\text{ if }i\ne j.
\end{cases}\tag6
$$
Substituting $(5)$ and $(6)$ into $(3)$, we get
$$
U_{123} = \text{CNOT}_{12}\otimes I_3 = \begin{bmatrix}
1&&&&&&&\\
&1&&&&&&\\
&&1&&&&&\\
&&&1&&&&\\
&&&&&&1&\\
&&&&&&&1\\
&&&&1&&&\\
&&&&&1&&
\end{bmatrix}\tag7
$$
and substituting $(5)$ and $(6)$ into $(3')$, we get
$$
U_{123}' = \text{CNOT}_{13}\otimes I_2 = \begin{bmatrix}
1&&&&&&&\\
&1&&&&&&\\
&&1&&&&&\\
&&&1&&&&\\
&&&&&1&&\\
&&&&1&&&\\
&&&&&&&1\\
&&&&&&1&
\end{bmatrix}.\tag{7'}
$$
Kronecker product
Finally, it is not hard to check that the above algorithm recovers the usual definition of the Kronecker product
$$
A\otimes B=\begin{bmatrix}a_{11}B&\dots&a_{1n}B\\&\dots&\\a_{n1}B&\dots&a_{nn}B\end{bmatrix}\tag8
$$
as a special case since in terms of matrix elements $(4)$ can be written as
$$
(A\otimes B)_{i_1i_2;k_1k_2} = A_{i_1;k_1}B_{i_2;k_2}\tag9
$$
where the compound indices $i_1i_2$ and $k_1k_2$ are to be interpreted as above.
