Notice firstly that $f(\operatorname{ker}f) = 0$ and $(\operatorname{coker}f)f = 0$. Use this to argue that $f = (\operatorname{im}f)f''(\operatorname{coim} f)$.
Once you've done that, we want to argue that $f''$ is an isomorphism. Thus $f = (\operatorname{im}f)(\operatorname{coim} f)$, justifying the factorization $f = me$, where $m$ is monic and $e$ epi. This is, in my view, where most of the work is.
After that, uniqueness (i.e. the universal property) more or less follows from the corresponding uniqueness and universal property statement for kernels, so I'll leave that to you.
Showing that $f''$ is an isomorphism requires some machinery, at least the way I know how (which is an amalgamation of Mac Lane's and Borceux's treatments): I feel a little strange screenshotting my own notes, but I don't know how to use AMS commutative diagrams, so
The bit on the next page is just noting that $mx = 0$, so since $m$ is monic, $x = 0$. To buy that this shows that $f''$ is an isomorphism, you have to know that in abelian categories,
- arrows that are both monic and epi are isomorphisms (not true in, say, Ring)
- that an arrow $g$ is monic (resp. epi) if and only if $gx = 0$ (resp. $xg = 0$) implies $x = 0$
- that pullbacks and pushouts exist and
- the pullback of an epi is an epi, and the pushout of a monic is monic.
