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.