I am trying to prove that $E'$ is an extension of $Q$ by $N'$
\begin{array} 00 &\longrightarrow & N & \overset{i_1} \longrightarrow & E & \overset{\pi_1} \longrightarrow& Q &\longrightarrow& 0 \\ & & \downarrow {\alpha}& &\downarrow {\beta} & & \downarrow {id}& &&\\ & & N' & \overset{i_2} \longrightarrow & E' & \overset{\pi_2} \longrightarrow & Q & & \end{array}
Where $E' = N' \oplus E \, /\ mod \, S$ where $S = \{(-\alpha(x),i_1(x)) \in N' \oplus E, \forall \, x \in N \}$
Now I think the answer is given here: Proposition 2
and can follow much of it, I am looking for a more general way using the snake lemma. What I have devised is
\begin{array} & & & 0 & & 0 & & 0 & & \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ 0 &\longrightarrow & ker (\alpha) & \overset{f} \longrightarrow & \ker (\beta) & \overset{g} \longrightarrow& ker (id) &\longrightarrow& 0 \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ 0 &\longrightarrow & N & \overset{i_1} \longrightarrow & E & \overset{\pi_1} \longrightarrow& Q &\longrightarrow& 0 \\ & & \downarrow {\alpha}& &\downarrow {\beta} & & \downarrow {id}& &&\\ &\ & N' & \overset{i_2} \longrightarrow & E' & \overset{\pi_2} \longrightarrow& Q & & \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ 0 &\longrightarrow & coker (\alpha) & \overset{j} \longrightarrow & coker (\beta) & \overset{k} \longrightarrow & coker(id) &\longrightarrow & 0 \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ & & 0 & & 0 & & 0 & & \\ \end{array} Which then gives $\delta: ker (id) \rightarrow coker (\alpha)$ so \begin{array} 00 &\longrightarrow & ker (\alpha) & \overset{f} \longrightarrow & \ker (\beta) & \overset{g} \longrightarrow& ker (id) & \overset{\delta} \longrightarrow& coker (\alpha) & \overset{j} \longrightarrow & coker (\beta) & \overset{k} \longrightarrow & coker(id) &\longrightarrow & 0 \end{array}
Which because $ker(id) = coker(id) = 0$ we get that:
$ker (\alpha) \simeq ker(\beta)$ and $coker (\alpha) \simeq coker(\beta)$
But am not sure in which way to proceed..