That is we start with an arbitrary extension $0 \to M \to e_2 \to L \to 0$ represented by the class of the image $\Phi_{e_2}:=\delta(id_L)$ with respect the delta-map in lower row in second diagram below and it's pullback extension $e_2$ in the upper row. Now we want determine that the extension $\overline{e_1}$ is represented by multiplication $a \cdot \Phi_{e_2} =: \Phi_{e_1}$.
We apply $Hom(L,-)$ to diagram
$$ \require{AMScd} \begin{CD} 0 @> >> M @> >> e_1 @>a^{-1} >> L @> >> 0\\ @VVV @VVV @VVV @VV\cdot{a}V \\ 0 @> >> M @> >> e_2 @> >> L @> >> 0 \end{CD} $$
and obtain
$$ \require{AMScd} \begin{CD} Hom(L, E) @> >> Hom(L,L) @>\delta >> Ext(L,M) @> >> \\ @VVV @VV\cdot{a}V @VVV \\ Hom(L,\overline{E}) @> >> Hom(L,L) @>\delta >> Ext(L,M) @> >> \end{CD} $$
That's a diagram of $k$-vector spaces. As you explaned in the answer the extension $e_1$ is forced to be the pullback of $e_2$: i.e. $e_1= a^*e_2$. $k$-linearity and commutativity of the maps imply $a \cdot \Phi_{e_2}=a \cdot \delta(id_L) = \delta(a \cdot id_L) = \Phi_{e_1}$. So $e_1=a e_2$. 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?