$\require{AMScd}$If in a regular category a square, whose sides $f,g,h,k$ are regular epis, is cartesian, is it cocartesian too? I would say yes, but I'm not sure that the proof below holds only under the hypothesis in italics; in particular I'm not convinced of the second paragraph. Do you think my (sketch of) proof is ok? Thank you
Consider this diagram, where $s_0,s_1$ is the kernel pair of $k$ (I drew only an arrow because I don't know how to draw two) and $r_0$ is the pullback of $s_0$ along $g$.
$$\begin{CD}
r@>{r_0}>> a @>f>> b\\
@VVjV @VVgV @VVhV\\
s@>{s_0,s_1}>>c @>k>> d
\end{CD}$$ One can prove that there is $r_1:r\to a$ such that $r_0,r_1$ is the kernel pair of $f$. This holds also inverting the roles of $r_0$ and $r_1$, and all kernel pairs (of $f$) are isomorphic, so if in the following diagram $r_0,r_1$ and $s_0 ,s_1$ are kernel pairs of $f$ and $k$, then the squares of sides $r_0,j,g,s_0$ and $r_1,j,g,s_1$ should be both cartesian. $$\begin{CD}
r@>{r_0,r_1}>> a @>f>> b\\
@VVjV @VVgV @VVhV\\
s@>{s_0,s_1}>>c @>k>> d
\end{CD}$$ Hence $j$ is a regular epi being the pullback of a regular epi.
Suppose that $u,v$ are two arrows such that $uf=vg$. Using that $j$ is epic one has that $vs_0=vs_1$, thus obtaining a unique $z$ such that $zk=v$; using that $f$ is epic follows $zh=u$ too, so $f,g,h,k$ is a cocartesian square.
Best Answer
Looks good to me! You can find a similar argument (albeit a much terser one) in an old answer here.
It seems like your main worry is with the claim that the kernel pair $r_0, r_1$ of $f : a \to b$ really is the pullback of the kernel pair $s_0, s_1$ of $k : c \to d$, so in the interest of providing a slightly less trivial answer, here's another way you might see that:
We meditate on the following commutative cube:
Notice that there's only one arrow $j : a \times_b a \to c \times_d c$ which simultaneously works for $r_0$ and for $r_1$. Indeed the maps $a \times_b a \to c \times_d c$ are in bijection with pairs of maps from $a \times_b a \to c$ whose coequalized by $k$, but both maps
$$ a \times_b a \overset{r_i}{\longrightarrow} a \overset{g}{\longrightarrow} c \overset{k}{\longrightarrow} d $$
are equal to
$$ a \times_b a \overset{r_i}{\longrightarrow} a \overset{f}{\longrightarrow} b \overset{h}{\longrightarrow} d $$
and the $r_i$ are coequalized by $f$, so these are the same arrow whether we choose $r_0$ or $r_1$.
Our task then, is to show that the left and back faces of this cube are pullbacks using the fact that the front, right, top, and bottom faces are all pullbacks. As you might expect, this is a "well known" generalization of the usual pasting lemma for pullbacks. But for completeness, let's prove it here:
Since the right and top faces are pullbacks, the usual pasting lemma tells us that the composite with legs $g r_1$, $k$, $r_0$, and $hf$ is a pullback. Of course, since the cube commutes, this tells us that the composite with legs $s_1 j$, $k$, $r_0$, and $kg$ is a pullback. But since the bottom square is a pullback, the converse of the pasting lemma tells us that the left face is a pullback (as desired).
A symmetric argument using the front and top faces, followed by the back and bottom faces shows that the back face is a pullback too.
From here, we can finish out your argument as you wrote it. Say we have maps $u : c \to z$ and $v : b \to z$
Since every reg epi (in particular $k$) coequalizes its kernel pair, we can get a (unique!) map $d \to z$ by showing that $u s_0 = u s_1$. Of course, since $j$ is a reg epi (as the pullback of a reg epi) it suffices to show that $u s_0 j = u s_1 j$, which (by commutativity) is the same as showing $u g r_0 = u g r_1$, and then $v f r_0 = v f r_1$. But this equation is true, since $f$ coequalizes $r_0$ and $r_1$!
This gives us a unique map $d \to z$ making the diagram commute, which shows that our original square was indeed a pushout.
I hope this helps ^_^