Let $(R,\mathfrak m)$ be a Noetherian local ring.
Let $F,G$ be finitely generated free $R$-modules and $f:F\to G$ be an $R$-linear map such that $f(F)\subseteq \mathfrak m G$. Let $X$ be a finitely generated $R$-module, and let $\sigma : 0\to F \to A_{\alpha} \to X \to 0$ be a short exact sequence i.e. $[\sigma]\in \text{Ext}^1_R(X,F)$.
We have a following push-out diagram with $[\beta] \in \text{Ext}^1_R(X,G)$.
$$
\require{AMScd}
\begin{CD}
\sigma : 0 @>>> F @>>> A_\alpha @>>> X @>>> 0 \\
@. @VV{f}V @VVV @| \\
\beta : 0 @>>> G @>>> A_\beta @>>> X @>>> 0
\end{CD}
$$
My question is:
Must it be necessarily true that $[\beta] \in \mathfrak m \text{Ext}^1_R(X,G)$?
Some thoughts:
The answer is affirmative if $F\cong G\cong R$. Indeed, in this case, $f:R\to R$ must be given by multiplication by some $x\in R$. Since $f(F)\subseteq \mathfrak m G$, so $x\in \mathfrak m$. Then in $[\beta]=x[\alpha]\in \text{Ext}^1_R(X,R)$, hence $[\beta]\in x\text{Ext}^1_R(X,R)\subseteq \mathfrak m \text{Ext}^1_R(X,G)$.
$\newcommand\m{\mathfrak{m}}\newcommand\Ext{\operatorname{Ext}}$ The class $[\beta]$ is just $\Ext_R^1(X,f)([\sigma])$, so it is enough to see that since $f:F\to G$ lands in $\m G$ then $$\Ext_R^1(X,f):\Ext_R^1(X,F)\to\Ext_R^1(X,G)$$ lands in $\m\Ext_R^1(X,G)$.
As $f$ factors as a composition $F\to \m G\hookrightarrow G$, it is in fact enough to check that the inclusion $mG\hookrightarrow G$ induces a map $$\Ext_R^1(X,\m G)\to\Ext_R^1(X,G)$$ that lands in $\m\Ext_R^1(X,G)$. You can compute this map by using a projective resolution of $X$ by finitely generated free modules. Do that.