I would like to prove that the connecting homomorphism $\delta \colon \mathrm{Hom}_{\mathcal{A}}(N,M_3) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$ from part (2) of Lemma 12.6.4 of the Stacks Project is indeed a group homomorphism.

In the proof, I would like to argue as follows:

By the definition of $\delta \colon \mathrm{Hom}_{\mathcal{A}}(N,M_3) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$, we know that $\delta(f)$ for some $f \in \mathrm{Hom}_{\mathcal{A}}(N,M_3)$ is simply the image of $[0 \to M_1 \xrightarrow{\alpha} M_2 \xrightarrow{\beta} M_3 \to 0] \in \mathrm{Ext}_{\mathcal{A}}(M_3,M_1)$ under the morphism $\mathrm{Ext}_{\mathcal{A}}(f,M_1) \colon \mathrm{Ext}_{\mathcal{A}}(M_3,M_1) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$.

In particular, if we are given $f,g \in \mathrm{Hom}_{\mathcal{A}}(N,M_3)$, then $\delta(f+g)$ equals the image of $[0 \to M_1 \xrightarrow{\alpha} M_2 \xrightarrow{\beta} M_3 \to 0] \in \mathrm{Ext}_{\mathcal{A}}(M_3,M_1)$ under the morphism $\mathrm{Ext}_{\mathcal{A}}(f+g,M_1) \colon \mathrm{Ext}_{\mathcal{A}}(M_3,M_1) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$.

However, it is well-known that $f+g \in \mathrm{Hom}_{\mathcal{A}}(N,M_3)$ may be described as the composition $$f+g \colon N \xrightarrow{\Delta} N \times N \xrightarrow{f \times g} M_3 \times M_3 \xrightarrow{\Sigma} M_3.$$ Moreover, by the definition of the Baer sum (cf. also this answer by Roland), $\delta(f) + \delta(g)$ is given by the image of $$[0 \to M_1 \times M_1 \to (M_2 \times_f N) \times (M_2 \times_g N) \to N \times N \to 0] \in \mathrm{Ext}_{\mathcal{A}}(N \times N,M_1 \times M_1)$$ ((here $M_2 \times_f N$ denotes the fibre product of $M_2 \xrightarrow{\beta} M_3 \xleftarrow{f} N$, and $M_2 \times_g N$ is defined similarly)) under $$\mathrm{Ext}_{\mathcal{A}} \Big( (N \times N, M_1 \times M_1) \xrightarrow{(\Delta^{\mathrm{opp}}, \Sigma)} (N,M_1) \Big) \colon \mathrm{Ext}_{\mathcal{A}}(N \times N, M_1 \times M_1) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1).$$ Further, I am quite sure that $$[0 \to M_1 \times M_1 \to (M_2 \times_f N) \times (M_2 \times_g N) \to N \times N \to 0] \in \mathrm{Ext}_{\mathcal{A}}(N \times N,M_1 \times M_1)$$ is the image of $$[0 \to M_1 \times M_1 \to M_2 \times M_2 \to M_3 \times M_3 \to 0] \in \mathrm{Ext}_{\mathcal{A}}(M_3 \times M_3,M_1 \times M_1)$$ under $\mathrm{Ext}_{\mathcal{A}}(f \times g, M_1 \times M_1) \colon \mathrm{Ext}_{\mathcal{A}}(M_3 \times M_3,M_1 \times M_1) \to \mathrm{Ext}_{\mathcal{A}}(N \times N,M_1 \times M_1)$.

But now there is one final step missing in order to conclude my argument relating $\delta(f+g)$ with $\delta(f)+\delta(g)$ in terms of the Ext bifunctor $\mathrm{Ext}_{\mathcal{A}} \colon \mathcal{A}^{\mathrm{opp}} \times \mathcal{A} \to \mathrm{Ab}$. I think the main question here is: How can I obtain the short exact sequence $0 \to M_1 \times M_1 \to M_2 \times M_2 \to M_3 \times M_3 \to 0$ via pullback/pushout when starting with the given short exact sequence $0 \to M_1 \to M_2 \to M_3 \to 0$?

Any help is kindly appreciated! In particular, if you have a reference for (or are willing to insert here) a proof of the fact that $\delta \colon \mathrm{Hom}_{\mathcal{A}}(N,M_3) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$ is a group homomorphism which does *not* follow the approach described above, then please feel free to let me know, too.

P.S.: I previously asked the same question on Mathematics StackExchange, but did not receive any comment/answer there.