Work in an abelian category. I'm aware that given an exact sequence, one can break it into short exact sequences like so:
I was wondering whether it was possible to do derive one of these short exact sequences from a single morphism alone: specifically, given just a single arrow $A \xrightarrow f B$, is it true that one can obtain a sequence $$ 0 \to \ker f \to A \to \operatorname{img} f \to 0 $$ which is short exact? I think this should be true (for example it's true in $\mathsf{Grp}$ and $\mathsf{Vect}_k$, among other categories) but I can't find a reference confirming this somehow.
Yes of course this is true, and in fact, this is said in the wikipedia article you linked !
Indeed, in the exact sequence involving the $A_i$'s, the $C_i$ do have a concrete description : $C_i=\ker(A_i\rightarrow A_{i+1})=\operatorname{im}(A_{i-1}\rightarrow A_i)$.
So if you just consider $f:A_1\rightarrow A_2$, you get a short exact sequence $$ 0\longrightarrow \ker f\longrightarrow A_1\longrightarrow \operatorname{im} f\longrightarrow 0$$
You even have a second one : $$ 0\longrightarrow \operatorname{im}f\longrightarrow A_2\longrightarrow\operatorname{coker} f\longrightarrow 0$$