We have that $\mathrm{Ext}^n(M, N) = \mathrm{Hom}(M, N[n])$, in other words, an element of $\mathrm{Ext}^n(M, N)$ is a map
$$
M \longrightarrow I_N[n]
$$
where $N \to I_N$ is an injective resolution. From this we obtain a distinguished triangle,
$$
M \longrightarrow I_N[n] \longrightarrow C \overset{+}\longrightarrow
$$
The homotopical invariant gives an extension of complexes
$$
0 \longrightarrow I_N[n] \longrightarrow C \longrightarrow M \longrightarrow 0 \
$$
which, plugin-in the beginning of the resolution of $N$, yields the classical $n$-extension
$$
0 \longrightarrow N \longrightarrow I^0_N \longrightarrow I^1_N \cdots  \longrightarrow I^{n-1}_N \longrightarrow M \longrightarrow 0
$$
as wanted