I thought I would add some support to/extend Michael’s answer by giving details of how to derive
\begin{equation*}
\mathrm{d}\tau^{2} = \left[ a^{2}\eta_{\rho\sigma}+a\left(b_{\sigma\lambda\rho}+b_{\rho\lambda\sigma}\right)x^{\lambda}+\cdots \right] \mathrm{d}x^{\rho}\mathrm{d}x^{\sigma}.
\end{equation*}
by using (3) in (2), see his answer.
Detailed Working.
I rewrote (2) and (3) as below
\begin{equation*}
\mathrm{d}\tau^{2}=\eta_{pq} \mathrm{d}\tilde{x}^{p}\mathrm{d}\tilde{x}^{q},\ (2^{\prime})
\end{equation*}
\begin{equation*}
\tilde{x}^{i}=ax^{i}+\frac{1}{2} b_{mn}^{i}x^{m}x^{n}+\cdots,\ (3^{\prime})
\end{equation*}
To use ($3^{\prime}$) in ($2^{\prime}$) we need an expression for $\mathrm{d}\tilde{x}^{i}$
\begin{equation*}
\mathrm{d}\tilde{x}^{i}=\frac{ \partial \tilde{ x }^i }{ \partial x^1 } \mathrm{d}x^{1}
+\frac{ \partial \tilde{ x }^i }{ \partial x^2 } \mathrm{d}x^{2}
+\frac{ \partial \tilde{ x }^i }{ \partial x^3 } \mathrm{d}x^{3}
+\frac{ \partial \tilde{ x }^i }{ \partial x^4 } \mathrm{d}x^{4}
\end{equation*}
Now, using ($3^{\prime}$) , but not the Einstein summation convention
\begin{align*}
\frac{ \partial \tilde{ x }^i }{ \partial x^1 } &= \frac{ \partial }{ \partial x^1 }
\left( ax^{i}+\frac{1}{2} \displaystyle \sum_{m,n=1}^4 b_{mn}^{i}x^{m}x^{n} \right) \\
&=a \delta_{i,1}+ \frac{ \partial }{ \partial x^1 } \frac{ 1}{ 2 }
\left( b_{11}^{i}x^1 x^1+ b_{12}^{i}x^1 x^2+ b_{13}^{i}x^1 x^3+ b_{14}^{i}x^1 x^4+
b_{21}^{i}x^2 x^1+ b_{31}^{i}x^3 x^1+ b_{41}^{i}x^4 x^1 \right) \\
&=a \delta_{i,1}+\frac{ 1}{ 2 }
\left( 2 b_{11}^{i}x^1 + 2 b_{12}^{i} x^2+ 2 b_{13}^{i}x^3+ 2 b_{14}^{i}x^4 \right )
\end{align*}
Where I dropped terms that would give zero when differentiated and then used the assumed symmetry of the $b's$.
I rewrite the above as
\begin{equation*}
\frac{ \partial \tilde{ x }^i }{ \partial x^1 } =a \delta_{i,1}+\displaystyle \sum_{j=1}^4 b_{1j}^{i}x^{j}
\end{equation*}
Similarly
\begin{equation*}
\frac{ \partial \tilde{ x }^i }{ \partial x^2 } =a \delta_{i,2}+\displaystyle \sum_{j=1}^4 b_{2j}^{i}x^{j}
\end{equation*}
\begin{equation*}
\frac{ \partial \tilde{ x }^i }{ \partial x^3 } =a \delta_{i,3}+\displaystyle \sum_{j=1}^4 b_{3j}^{i}x^{j}
\end{equation*}
\begin{equation*}
\frac{ \partial \tilde{ x }^i }{ \partial x^4 } =a \delta_{i,4}+\displaystyle \sum_{j=1}^4 b_{4j}^{i}x^{j}
\end{equation*}
\begin{equation*}
\mathrm{d}\tilde{x}^{i}=\frac{ \partial \tilde{ x }^i }{ \partial x^1 } \mathrm{d}x^{1}
+\frac{ \partial \tilde{ x }^i }{ \partial x^2 } \mathrm{d}x^{2}
+\frac{ \partial \tilde{ x }^i }{ \partial x^3 } \mathrm{d}x^{3}
+\frac{ \partial \tilde{ x }^i }{ \partial x^4 } \mathrm{d}x^{4} \\
\end{equation*}
\begin{align*}
\mathrm{d}\tilde{x}^{i}=&+a(\delta_{i,1} \mathrm{d}x^{1} + \delta_{i,2} \mathrm{d}x^{2}+\delta_{i,3} \mathrm{d}x^{3} + \delta_{i,4} \mathrm{d}x^{4}) \\
&+ \displaystyle \sum_{j=1}^4 b_{1j}^{i}x^{j} \mathrm{d}x^{1}
+\displaystyle \sum_{j=1}^4 b_{2j}^{i}x^{j} \mathrm{d}x^{2}
+\displaystyle \sum_{j=1}^4 b_{3j}^{i}x^{j} \mathrm{d}x^{3}
+\displaystyle \sum_{j=1}^4 b_{4j}^{i}x^{j} \mathrm{d}x^{4}
\\
=&+a\mathrm{d}x^{i} \\
&+\displaystyle \sum_{j=1}^4
( b_{1j}^{i}x^{j} \mathrm{d}x^{1}+ b_{2j}^{i}x^{j} \mathrm{d}x^{2}
+ b_{3j}^{i}x^{j} \mathrm{d}x^{3}+ b_{4j}^{i}x^{j} \mathrm{d}x^{4} ) x^j \\
\end{align*}
\begin{equation*}
\mathrm{d}\tilde{x}^{i}=a\mathrm{d}x^{i} + \displaystyle \sum_{k,j=1}^4
b_{kj}^{i}x^{j} \mathrm{d}x^{k}
\end{equation*}
So using this, in
\begin{equation*}
\mathrm{d}\tau^{2}=\eta_{pq} \mathrm{d}\tilde{x}^{p}\mathrm{d}\tilde{x}^{q},\ (2^{\prime})
\end{equation*}
\begin{align*}
\mathrm{d}\tau^{2}= \displaystyle \sum_{p,q=1}^4
\eta_{pq}(a \mathrm{d}x^{p}&+ \displaystyle \sum_{k,j=1}^4 b_{kj}^{p}x^{j} \mathrm{d}x^{k}) \\
( a\mathrm{d}x^{q}&+ \displaystyle \sum_{r,s=1}^4 b_{rs}^{q}x^{s} \mathrm{d}x^{r} )
\end{align*}
We have four terms, as the $x^j$ are small one is dropped, the first term appears in Michaels result and is given by
\begin{equation*}
\displaystyle \sum_{p,q=1}^4 a^2 \eta_{pq} \mathrm{d}x^{p} \mathrm{d}x^{q}
\end{equation*}
or, using his notation and the Einstein summation convention
\begin{equation*}
a^2 \eta_{\rho \sigma} \mathrm{d}x^{\rho} \mathrm{d}x^{\sigma}
\end{equation*}
Let
\begin{equation*}
\mathrm{d}\tau^{2} = T_1+T_2+T_3
\end{equation*}
Where
\begin{align*}
T_1 &= \displaystyle \sum_{p,q=1}^4 \eta_{pq} a^2 \mathrm{d}x^{p} \mathrm{d}x^{q}= a^2 \eta_{\rho \sigma} \mathrm{d}x^{\rho} \mathrm{d}x^{\sigma}\\
T_2 &= \displaystyle \sum_{p,q=1}^4 \eta_{pq}~ a~ \mathrm{d}x^{p}
\displaystyle \sum_{r,s=1}^4 b_{rs}^{q}~x^s\mathrm{d}x^{r} \\
T_3 &= \displaystyle \sum_{p,q=1}^4 \eta_{pq} \displaystyle \sum_{k,j=1}^4 b_{kj}^{p}~x^j\mathrm{d}x^{k}
~a~ \mathrm{d}x^{q}
\end{align*}
Both $T_2 \text{and}~T_3$ may be re-expressed in quite similar ways.
$T_2$ may be re-written in the form
\begin{equation*}
T_2 = a~\displaystyle \sum_{p,q=1}^4\displaystyle \sum_{r,s=1}^4 \eta_{pq}~ \mathrm{d}x^{p} ~b_{rs}^{q}~x^s\mathrm{d}x^{r}
\end{equation*}
Or, as
\begin{equation*}
T_2 = a~\displaystyle \sum_{p=1}^4 \mathrm{d}x^{p}
\displaystyle \sum_{s=1}^4 x^s
\displaystyle \sum_{r=1}^4 \mathrm{d}x^{r}
\displaystyle \sum_{q=1}^4 \eta_{pq}~b_{rs}^{q}
\end{equation*}
Define
\begin{equation*}
b_{\alpha \beta \gamma} =\displaystyle \sum_{t=1}^4 \eta_{\alpha t}~b_{\gamma \beta}^{t}
\end{equation*}
then
\begin{equation*}
\displaystyle \sum_{q=1}^4 \eta_{pq}~b_{rs}^{q}=\displaystyle \sum_{t=1}^4 \eta_{pt}~b_{rs}^{t}
=b_{psr}
\end{equation*}
\begin{equation*}
T_2 = a~\displaystyle \sum_{p=1}^4 \mathrm{d}x^{p}
\displaystyle \sum_{s=1}^4 x^s
\displaystyle \sum_{r=1}^4 \mathrm{d}x^{r} ~b_{psr}
\end{equation*}
Re-label ‘$r$’ as ‘$q$’
\begin{equation*}
T_2 = a~\displaystyle \sum_{p,q=1}^4 ~\displaystyle \sum_{s=1}^4~b_{psq}~ x^s \mathrm{d}x^{p}~\mathrm{d}x^{q}
\end{equation*}
Using Michael’s notation so $p,q,s \to \rho , \sigma , \lambda$ , and the Einstein summation convention
\begin{equation*}
T_2 = a~b_{\rho\lambda\sigma}~x^{\lambda}~~\mathrm{d}x^{\rho}\mathrm{d}x^{\sigma}
\end{equation*}
$T_3$ may be re-written as
\begin{equation*}
T_3 = a~\displaystyle \sum_{q=1}^4 \mathrm{d}x^{q}
\displaystyle \sum_{j=1}^4 x^j
\displaystyle \sum_{k=1}^4 \mathrm{d}x^{k}
\displaystyle \sum_{p=1}^4 \eta_{pq}~b_{kj}^{p}
\end{equation*}
There is a slight wrinkle to handle. Using the symmetry of $\eta_{pq}$, we note that
\begin{equation*}
\displaystyle \sum_{p=1}^4 \eta_{pq}~b_{kj}^{p}=\displaystyle \sum_{p=1}^4 \eta_{qp}~b_{kj}^{p}
\end{equation*}
Then using the definition of $b_{\alpha \beta \gamma}$
\begin{equation*}
\displaystyle \sum_{p=1}^4 \eta_{qp}~b_{kj}^{p}=
\displaystyle \sum_{t=1}^4 \eta_{qt}~b_{kj}^{t}=b_{qjk}
\end{equation*}
So,
\begin{equation*}
T_3 = a~\displaystyle \sum_{q=1}^4 \mathrm{d}x^{q}
\displaystyle \sum_{j=1}^4 x^j
\displaystyle \sum_{k=1}^4 \mathrm{d}x^{k}~b_{qjk}
\end{equation*}
Re-label ‘$k$’ as ‘$p$’
\begin{equation*}
T_3 = a~\displaystyle \sum_{p,q=1}^4 \displaystyle \sum_{j=1}^4 ~b_{qjp}~x^j
\mathrm{d}x^{p}~ \mathrm{d}x^{q}
\end{equation*}
Using Michael’s notation so $p,q,j \to \rho , \sigma , \lambda$ , and the Einstein summation convention
\begin{equation*}
T_3 = a~b_{\sigma \lambda \rho }~x^{\lambda}~~\mathrm{d}x^{\rho}\mathrm{d}x^{\sigma}
\end{equation*}
Collecting the results for $T_1,~T_2,~T_3$
\begin{equation*}
\mathrm{d}\tau^{2} = T_1+T_3+T_2~~~~~~~~~~~~,\text{meant}
\end{equation*}
\begin{equation*}
\mathrm{d}\tau^{2} = \left[ a^{2}\eta_{\rho\sigma}+a\left(b_{\sigma\lambda\rho}+b_{\rho\lambda\sigma}\right)x^{\lambda} \right] \mathrm{d}x^{\rho}\mathrm{d}x^{\sigma}
\end{equation*}
Which confirms one of Michael’s stated results.
Pleas note that the $b$'s with three subscripts, are not defined in Michael's answer.
