That is we start with an arbitrary extension $0 \to M \to E \to L \to 0$ represented by the class of the image $\Phi_E:=\delta(id_L)$ with respect the delta-map in upper row in second diagram below and we want to determine which extension $\overline{E}$ is represented by $a \cdot \Phi_E =: \Phi_{\overline{E}}$.


We apply $Hom(L,-)$ to diagram

$$
\require{AMScd}
\begin{CD}
0 @>  >> M @>  >> E @> >>  L @> >> 0\\
@VVV  @VVV  @VVV  @VV\cdot{a}V \\
0 @>  >> M @>  >> \overline{E} @> >>  L @> >> 0 
\end{CD}
$$

and obtain


$$
\require{AMScd}
\begin{CD}
Hom(L, E) @>  >> Hom(L,L) @>\delta  >> Ext(L,M) @> >>  \\
@VVV  @VV\cdot{a}V  @VV\cdot{a}V  \\
Hom(L,\overline{E}) @>  >> Hom(L,L) @>\delta  >> Ext(L,M) @> >> 
\end{CD}
$$

That's a diagram of $k$-vector spaces. As you explaned in the answer the extension $E$ is forced to be the pullback of $\overline{E}$: i.e. $E= a^*\overline{E}$. This implies that $\overline{E}= (a^{-1})^*E$, thus $a \cdot \Phi_E = \Phi_{(a^{-1})^*E}$. Is this the correct result of the $k^*$ action by scalar multiplication on $Ext(L,M)$? Or do I have somewhere implemented your hints on my question 1) in wrong way?