For the sake of keeping this post self-contained, I should include a description of how to characterize $\operatorname{Ext}^n$ as $n$-extensions as outlined in this section on Wikipedia.
For two objects $A$ and $B$ in an abelian category, define an $n$-extension of $B$ by $A$, as an exact sequence
$$
0
\to A
\to X_1
\to \dotsb
\to X_n
\to B
\to 0
\,.
$$
Then we want to say two $n$-extensions are equivalent if there are maps $\{f_1, \dotsc,f_n\}$ such that every square in this diagram commutes:
$$
\require{AMScd}
\begin{CD}
0 @>>> A @>>> X_1 @>>> \dotsb @>>> X_n @>>> B @>>> 0\\
@.@V{\mathrm{id}}VV @V{f_1}VV @. @V{f_n}VV @V{\mathrm{id}}VV \\
0 @>>> A @>>> Y_1 @>>> \dotsb @>>> Y_n @>>> B @>>> 0\\
\end{CD}
$$
This relation is clearly reflexive and transitive. Showing that it is symmetric, and can indeed be regarded as an honest equivalence relation takes some work (look for a proof of the short-five lemma to help). Under this relation, there is a set bijection between equivalence classes of $n$-extensions and $\operatorname{Ext}^n(B,A)$.
We can strengthen this result by imposing a group structure on this set of equivalence classes and saying that it and $\operatorname{Ext}^n(B,A)$ are isomorphic as groups. Let $X_n \times_B Y_n$ be the pullback of $X_n$ and $Y_n$ over $B$, and let $X_1 +_A Y_1$ be the pushout of $X_1$ and $Y_1$ over $A$. Define the Baer sum of the two $n$-extensions above as
$$
0
\to A
\to X_1 +_A Y_1
\to X_2 \oplus Y_2
\to \dotsb
\to X_{n-1} \oplus Y_{n-1}
\to X_n \times_B Y_n
\to B
\to 0
\,.
$$
This is another $n$-extension of $B$ by $A$. The Baer sum imposes a group structure on the set of equivalence classes of $n$-extensions, the identity element being the class of split $n$-extensions with a canonical representative
$$
X_i =
\underbrace{A \oplus \dotsb \oplus A}_{\binom n i \text{ copies}}
\oplus
\underbrace{B \oplus \dotsb \oplus B}_{\binom n {n-i+1} \text{ copies}}
\;,
$$
making it isomorphic to the group $\operatorname{Ext}^n(B,A)$.