An epimorphism is called split if it has a section (a right inverse).
An epimorphism is called effective if it has a kernel pair and is the coequalizer of its kernel pair.
The ncatlab implies here that every split epimorphism is effective, but I can't find a proof anywhere, nor have I had success in coming up with a proof or counterexample.
There are two parts to this. First, is it true that every split epimorphism has a kernel pair? I haven't managed to come up with an obvious construction of its kernel pair; what strikes me as the natural choice doesn't seem to work. (My inclination would be to try, if $f$ is a split epi with section $s$, the pair $(s \circ f, 1)$, because I think $f$ is definitely the coequalizer of this - but that doesn't seem to work.)
Second: if every split epimorphism does have a kernel pair, then is it further true that this kernel pair must have a coequalizer?
While it definitely suggests it, I don't think the nLab page means that every split epimorphisms is an effective epimorphism but only every split epimorphism is a coequalizer of its kernel pair if it has one. For example, on the split epimorphism page it does not state that a split epimorphism is an effective epimorphism, only that it is a regular epimorphism. There is also this note (emphasis added):
Let $r : A \to B$ and $r\circ s = id$. Assume the kernel pair $(R,p,q)$ of $r$ exists, i.e. $$\require{AMScd}\begin{CD}R@>q>> A \\@VpVV@VVrV\\A@>>r> B\end{CD}$$ is a pullback square. We want to show that for any $h:A \to Z$ such that $h\circ p=h\circ q$, there exists a unique $t : B \to Z$ such that $t\circ r = h$. We can immediately see that $t$ can only be $h\circ s$, so it remains to show that $h\circ s\circ r = h$. Since $r\circ s\circ r = r$, we have that there exists a unique $k: B \to R$ such that $p\circ k = s\circ r$ and $q\circ k = id$. This means that $h\circ s \circ r = h\circ p \circ k = h\circ q \circ k = h$ as needed. $\square$