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?