Here is my own try for an answer, assuming further that $\mathcal{A}$ has no torsion. I wish to show that the Baer sum $nE = E + \ldots + E$ corresponds to replacing $\eta$ by $n\eta$.

My idea is this, let $P$ denote the pullback of $E\oplus\ldots \oplus E$ (the $n$-fold direct sum) by the $n$-fold diagonal $B\to B\oplus\ldots \oplus B$. It is an extension of $B$ by the $n$-fold direct sum of $n$-copies of $A$.

Its elements correspond to $n$-tuples of elements of $E$ which have the same image in $B$. There is an obvious trace map from $P$ to $E_n$, which is defined to be the object $E$ with the extension structure corresponding to replacing $\eta$ by $n\eta$.

Since the trace map has a section (assuming multiplication by $n$ is an isomorphism, as $\mathcal{A}$ is torsion free), the trace map is an epimorphism. Its kernel consists of $n$-tuples $(e_1,\ldots,e_n)$ which sum to $0$ in $E_n$ and have the same image in $B$, denote this image by $b$, then $nb=0$, again under our torsion freeness assumption, $b=0$.

Therefore all the $e_i$ are in $A$ and sum to 0, hence the kernel of the projection is the kernel of the $n$-fold summation on $A$, hence $E_n$ is isomorphic to $nE$, by definition.

We can see that this arguement wouldn't work for $n=0$, which is indeed an exceptional case.

Still unclear what happens if $E$ or $B$ are allowed to have $n$-torsion.

Here is my own try for an answer, assuming further that multiplication by $n$ on $\mathcal{A}$ is invertible for all $n\neq 0$, and specializing to a category of modules for simplicity. I wish to show that the Baer sum $nE = E + \ldots + E$ corresponds to replacing $\eta$ by $n\eta$.

My idea is this, let $P$ denote the pullback of $E\oplus\ldots \oplus E$ (the $n$-fold direct sum) by the $n$-fold diagonal $B\to B\oplus\ldots \oplus B$. It is an extension of $B$ by the $n$-fold direct sum of $n$-copies of $A$.

Its elements correspond to $n$-tuples of elements of $E$ which have the same image in $B$. There is an obvious trace map from $P$ to $E_n$, which is defined to be the object $E$ with the extension structure corresponding to replacing $\eta$ by $n\eta$.

Since the trace map has a section (assuming multiplication by $n$ is an isomorphism, as $\mathcal{A}$ is torsion free), the trace map is an epimorphism. Its kernel consists of $n$-tuples $(e_1,\ldots,e_n)$ which sum to $0$ in $E_n$ and have the same image in $B$, denote this image by $b$, then $nb=0$, again under our torsion freeness assumption, $b=0$.

Therefore all the $e_i$ are in $A$ and sum to 0, hence the kernel of the projection is the kernel of the $n$-fold summation on $A$, hence $E_n$ is isomorphic to $nE$, by definition.

We can see that this arguement wouldn't work for $n=0$, which is indeed an exceptional case.

Still unclear what happens if $E$ or $B$ are allowed to have $n$-torsion.