Let $\mathcal{A}$ be an abelian category and let $$ 0 \rightarrow E \rightarrow F \rightarrow G \rightarrow 0 $$ be a short exact sequence. Then in $D(\mathcal{A})$, the derived category of $\mathcal{A}$ we have a distinguished triangle $$ E \rightarrow F \rightarrow G \rightarrow E[1]. $$

Moreover, the above short exact sequence determines an element of Ext$^1(G,E)$ and therefore a map $G[-1] \rightarrow E$ in $D(\mathcal{A})$. The axiom TR2 tells us that the cone of this map is exactly $F$.

I would like to understand if one can generalise this picture.

Consider a morphism $f: A \rightarrow B$ in $\mathcal{A}$ such that both Ker$(f)$ and Coker$(f)$ are non-zero. $f$ can be also thought of as a morphism in $D(\mathcal{A})$. What is its cone then? Do we have a distinguished triangle $$ A \rightarrow B \rightarrow \textrm{Coker}(f) \oplus \textrm{Ker}(f)[1] \rightarrow A[1]? $$

On the other hand, we have the following exact sequence in $\mathcal{A}$ $$ 0 \rightarrow \textrm{Ker}(f) \rightarrow A \rightarrow B \rightarrow \textrm{Coker}(f) \rightarrow 0. $$ It corresponds to an element in $\textrm{Ext}^2(\textrm{Coker}(f), \textrm{Ker}(f))$ and hence to a map $\textrm{Coker}(f)[-2] \rightarrow \textrm{Ker}(f)$ in $D(\mathcal{A})$. Can one write down the cone of this map using objects $A$ and $B$? My only guess is that it would be $A \oplus B[-1]$, but I don't know whether it is correct or how to prove it.