Baer Sum notation requires clearence.

I guess the author meant the following : let $\operatorname{Ext}^1(A,B)$ be the set of extension of $A$ by $B$. Then $\operatorname{Ext}^1(A,B)$ is a covariant functor in $B$ and a contravariant functor in $A$.

So, if $f:A'\rightarrow A$ and $E\in\operatorname{Ext}^1(A,B)$, you can take its image in $\operatorname{Ext}^1(A,B)$, this image will be denoted by $E\circ f\in\operatorname{Ext}^1(A',B)$. Namely, $E\circ f$ is the pull-back $E\circ f=A'\times_A E$ : $$\require{AMScd} \begin{CD} 0@>>>B@>>>E\circ f@>>>A'@>>>0\\ @.@|@VVV@VfVV@.\\ 0@>>>B@>>>E@>>>A@>>>0 \end{CD}$$

And similarly, if $g:B\rightarrow B'$ is any morphism, write $g\circ E\in\operatorname{Ext}^1(A,B')$ for the corresponding extension. It is dually the pushout : $$\require{AMScd} \begin{CD} 0@>>>B@>>>E@>>>A@>>>0\\ @.@VgVV@VVV@|@.\\ 0@>>>B'@>>>g\circ E@>>>A@>>>0 \end{CD}$$

Now you have an extension $E\oplus F\in\operatorname{Ext}^1(A\oplus A,B\oplus B)$. Just take $\nabla_B\circ(E\oplus F)\circ\Delta_A$. In other words, the Baer sum of $E$ and $F$ is exactly your quotient $X/N$.