Fix a additive category $\mathsf C$ (that is to say, $\mathsf C$ admits an additive structure and has a zero object all biproducts). Given four morphisms in $\mathsf C$ as follows:
$$\require{AMScd} \begin{CD} x @>{\psi}>> a\\ @V{\phi}VV @VV{f}V \\ b @>>{g}> c \end{CD}$$
Suppose $(\psi,\phi)$ is a pullback of $(f,g)$ and that $f\equiv\text{coker}(\ker f)$. Can we deduce that $\phi$ is also epic?
Here's a counterexample. Consider $R=\mathbb{Z}[t]$ as a graded ring with $|t|=1$, and let $D$ be the category of graded $R$-modules. Define the objects $a=R$, $k=(t^2)\subset R$, $c=R/(t^2)$, and $b=(t)/(t^2)\subset c$ in $D$. Let $C$ be the full subcategory of $D$ spanned by finite direct sums of the objects $a$, $k$, $c$, and $b$.
Note that there is an obvious inclusion map $k\to a$ whose cokernel is a regular epimorphism $f:a\to c$. Let $g:b\to c$ be the inclusion map. Now suppose $x$ is any object of $C$ and you have maps $\psi:x\to a$ and $\phi:x\to b$ such that $f\psi=g\phi$. Note that there are no nonzero maps from any of $a$, $k$, or $c$ to $b$. So $\phi$ must vanish on all the summands of $a$, $k$ and $c$ in the object $x$. On the other hand, there are no nonzero maps from $b$ to $a$, so $\psi$ must vanish on all the summands of $b$ in the object $x$. Since $g$ is injective, this means that $\phi$ must also vanish on all the summands of $b$ in $x$. Thus $\phi=0$.
From the paragraph above, it is now easy to see that $x=k$ with $\psi:x\to a$ the inclusion map and $\phi:x\to b$ the $0$ map is a pullback of $a$ and $b$ over $c$ in $C$. The map $\phi$ for this pullback is not epic.
More generally, you can get a similar counterexample whenever you have an abelian category with a short exact sequence $0\to k\to a\to c\to 0$ and a subobject $b\to c$ such that there are no nonzero maps from $k$, $a$, or $c$ to $b$ and no nonzero maps from $b$ to $a$.