Let $\mathscr{A}$ be an Abelian category, and let $$A\xrightarrow{f} B\xrightarrow{g} C\to 0$$ be an exact sequence in $\mathscr{A}$, and let $h:X\to B$ be any morphism, which gives us a sequence $$A\xrightarrow{f}B\twoheadrightarrow\mathrm{coker\,}h\to\mathrm{coker\,}gh\to 0$$ where the map $\mathrm{coker\,}h\to\mathrm{coker\,}gh$ is given by universal properties. If we collapse the first two maps, I claim that the resulting sequence $$A\to\mathrm{coker\,}h\to\mathrm{coker\,}gh\to 0$$ is exact.
Using relatively elementary methods, I've managed to prove that this is a complex, and that the map $\mathrm{coker\,}h\to\mathrm{coker\,}gh$ is surjective. But when I try to prove that $$\mathrm{im}\left(A\xrightarrow{f}B\twoheadrightarrow\mathrm{coker\,}h\right) = \ker\left(\mathrm{coker\,}\to\mathrm{coker\,}gh\right)$$ I can't seem to line up any of the arrows to use the universal properties of the kernel and cokernel.
This isn't too hard to prove with spectral sequences. Even if you are unfamiliar with them, maybe they'll still be inspirational.
Consider the bicomplex
$$ \begin{matrix} 0 &\to& X &\to& X \\ \downarrow & & \downarrow & & \downarrow & & \\ A &\to& B &\to& C \\ \downarrow & & \downarrow & & \downarrow & & \\ A &\to& \operatorname{coker}{h} &\to& \operatorname{coker}(gh) \end{matrix} $$
You can compute the homology of its total complex in two different ways. If we start by taking the horizontal homology, the next page of the spectral sequence is
$$ \begin{matrix} 0 & 0 & 0 \\ \bullet & 0 & 0 \\ \bullet & U & V \end{matrix} $$ where $\bullet$ indicates groups we don't care about. This implies the bottom two homology groups are $U$ and $V$.
If we compute by taking vertical homology first, the next page of the spectral sequence is $$ \begin{matrix} 0 & \bullet & \bullet \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} $$ from which we can determine the bottom two homology groups are both zero.