Consider the diagram below:
The pushout of the corner is $C=(Y\sqcup Z)/\sim$ where $$Y\sqcup Z=\{(y,\star):y\in Y\}\cup\{(\star,z):z\in Z\}$$ and $\sim$ is the equivalence relation generated by $$R=\{((s(x),\star),(\star,t(x))):x\in X\}$$ together with projection maps $p_1:Y\to C,y\mapsto[(y,\star)]$ and $p_2:Z\to C,z\mapsto [(\star,z)]$. It is clear that if $\cdot$ in the diagram above is replaced by $C$, then the diagram commutes.
Suppose $f_1:Y\to A$ and $f_2:Z\to A$ be arrows such that if $\cdot$ is replaced with $A$ then the diagram also commutes. To prove that $C$ is a colimit, we must prove that there is a unique map $\phi:C\to A$ with $\phi p_i=f_i$.
Define $$\phi([c]) = \begin{cases} f_1(y) & \text{ if } c=(y,\star) \\ f_2(z) & \text { if } c=(\star,z)\end{cases}$$
As long as this map is well defined, it is the unique map that makes the triangles commute, by construction.
I'm confused about the proof that it's well-defined. The idea should be the same as in this answer but in our case $\phi$ has definition by cases, which confuses me. I'm not even sure how to define the equivalence relation referred to in the answer cited because the elements of $Y\sqcup Z$ are not of the same form, so I'm not sure how to say $w\approx w'$ iff $f_1(...)=f_2(...)$ -- each of $w,w'$ is either of the form $(y,\star)$ or $(\star,z)$. In particular, this is a question about notation - I don't know how to organize it properly.
Define $\tilde \phi : Y \sqcup Z \to A$ by the corresponding rules $\tilde \phi((y, \star)) = f_1(y)$ and $\tilde \phi((\star, z)) = f_2(z)$. Then check that $\{ (a, b) \in (Y \sqcup Z) \times (Y \sqcup Z) \mid \tilde \phi(a) = \tilde \phi(b) \}$ is an equivalence relation on $Y \sqcup Z$, and that it contains all elements of $R$. It then follows that this relation contains the equivalence relation generated by $R$, which amounts to exactly the same thing that you wanted to prove.