Let us consider the following constructions in the category of $R-$modules, for some ring $R$.
Given a short exact sequence
$$ \mathcal{S}: \quad 0 \to A \overset{\alpha}{\to} B \overset{\beta}{\to} C \to 0$$
and a map $\phi : A \to Z$, we have the pushout
$$B \sqcup_A Z = B \oplus Z \ / \ (\alpha(a),0) \sim (0,\phi(a)),$$
that fits into the following short exact sequence:
$$ 0 \to Z \overset{u}{\to} B \sqcup_A Z \overset{v}{\to} C\to 0.$$
Similarly, given a map $\psi:Z \to C$, we can form the pullback $$ Z \times_C B = \{ (z,b) \in Z \oplus B \ \vert \ \psi(z)=\beta(b)\},$$
that fits into the following short exact sequence:
$$ 0 \to A \overset{r}{\to} Z \times_C B \overset{s}{\to} Z \to 0.$$
We can even combine these constructions and obtain the short exact sequences
$$ 0 \to A \to (B \sqcup_A Z )\times_C B \to B \sqcup_A Z \to 0,$$ $$ 0 \to Z \times_C B \to B \sqcup_A (Z \times_C B) \to C \to 0.$$
More explicit descriptions of these modules are as follows: \begin{alignat*}{3} &(B \sqcup_A Z )\times_C B &&= &&\left\{ \left([(b,z)],b'\right) \in \left( B \oplus Z \ / \ (\alpha(a),0) \sim (0, \phi(a)) \right) \oplus B \ \middle\vert \ v([(b,z)]) = \beta(b')\right\},\\ &B \sqcup_A( Z \times_C B) &&= &&\left\{ (b',(z,b)) \in B\oplus (Z \oplus B) \ \middle\vert \ \psi(z) = \beta(b) \right\} \ / \ (\alpha(a),0,0) \sim (0,r(a)). \end{alignat*}
All maps are the obvious ones, in particular $v([(b,z)]) = \beta(b)$ and $r(a) = (0,\alpha(a))$.
If $\phi$ and $\psi$ fit into a short exact sequence $$ \mathcal{T}: \quad 0 \to A \overset{\phi}{\to} Z \overset{\psi}{\to} C \to 0,$$
then we can also form the Baer sum
$$\mathcal{S} \boxplus \mathcal{T}: \quad 0 \to A \to B\boxplus Z \to C \to 0,$$
where $B \boxplus Z$ is defined as
$$ B\boxplus Z:=\{ (b,z) \in B\oplus Z \ \vert \ \beta(b)=\psi(z) \} \ / \ (\alpha(a),0) \sim (0, \phi(a)).$$
The three constructions have a lot in common, and I am really puzzled as to their relationship. I cannot even figure out if all three $R-$modules are isomorphic or if they are completely different things, and, if they are not, whether there is a nice way to relate them.
In general, in an abelian category, you can take pullbacks and pushouts of any two maps; you don't need injectivity or surjectivity.
Given two maps $f\colon A\to B$ and $g\colon A\to C$, the pushout is the cokernel $D$ of the map $\binom{f}{g}\colon A\to B\oplus C$ (together with the pair of maps $B\to D$ and $C\to D$). The universal property of the pushout corresponds directly to the universal property of the cokernel. Similarly, the pullback can be constructed as a kernel.
Next, sticking to the pair of maps $f\colon A\to B$ and $g\colon A\to C$, if the map $g$ factors through $f$, so $g=hf$ for some $h$, then the pushout is isomorphic to $\mathrm{Cok}(f)\oplus C$. For, we have an isomorphism $\phi=\begin{pmatrix}1&0\\-h&1\end{pmatrix}$ of $B\oplus C$, composition yields $\phi\binom fg=\binom f0$, and the cokernel of $\binom f0$ is of course $\mathrm{Cok}(f)\oplus C$.
Applying these observations to your situation, the map $\alpha\colon A\to B$ factors through the map $r\colon A\to Z\times_CB$ coming from the pullback, and hence the pushout of these two maps is $\mathrm{Cok}(\alpha)\oplus(Z\times_CB)=C\oplus(Z\times_CB)$. Thus the short exact sequence you write down $$ 0 \to Z\times_CB \to B\sqcup_A(Z\times_CB) \to C \to 0 $$ is actually split exact. Similarly for the other such sequence.
For the Baer sum, we start with two short exact sequences $$ 0 \to A \to B \to C \to 0 \quad\textrm{and}\quad 0 \to A \to B' \to C \to 0, $$ form their direct sum $$ 0 \to A^2 \to B\oplus B' \to C^2 \to 0, $$ and then take the pushout along the codiagonal $(1,1)\colon A^2\to A$ and the pullback along the diagonal $\binom11\colon C\to C^2$ (these two constructions commute) to obtain a new short exact sequence $$ 0 \to A \to B'' \to C \to 0, $$ which is the Baer sum of the starting pair. Really we are doing this on equivalence classes of short exact sequences, but everything is OK.)