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 $$ I_M \longrightarrow N[n] $$ where $M \to I_M$ is an injective resolution, in other words, induces a sequence $$ 0 \to M \to I_1 \to \cdots I_n \to N \to 0 $$ which corresponds to the classical $n$-extensions. The last map is a surjection because the derived category map goes trough the homology of the complex.

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

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 $$ I_M \longrightarrow N[n] $$ where $M \to I_M$ is an injective resolution, in other words, induces a sequence $$ 0 \to M \to I_0 \to \cdots I_n \to N \to 0 $$ which corresponds to the classical $n$-extensions. The last map is a surjection because the derived category map goes trough the homology of the complex.

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 $$ I_M \longrightarrow N[n] $$ where $M \to I_M$ is an injective resolution, in other words, induces a sequence $$ 0 \to M \to I_1 \to \cdots I_n \to N \to 0 $$ which corresponds to the classical $n$-extensions. The last map is a surjection because the derived category map goes trough the homology of the complex.