This is related to Weibel, Homological Algebra, Thm 3.4.3 proof. I am only considering in R module category. The goal is to check that well-definedness map from equivalence class of extensions between $A,B$ to $Ext^1(A,B)$.
Given a resolution $0\to M\to P\to A\to 0$ of $A$, with $P$ projective and $x\in Ext^1(A,B)$, one can obtain a map $f:M\to B$ by application of $Hom(-,B)$ functor. This is related to exact sequence part $Hom(P,B)\to Hom(M,B)\to Ext^1(A,B)\to 0$.
Let $X_f$ be pushout of $M\to P$ and $M\to B$. Suppose $f'$ is another element with coboundary image $x\in Ext^1(A,B)$ which implies $f'-f$ lifts to $g\in Hom(P,B)$. Then I have another pushout $X_{f'}$. Now I have to check $X_{f'}\cong X_f$.
I can consider the following. Consider $B\oplus P\to B\oplus P$ automorphism by $(b,p)\to (b-g(p),p)$ and clearly the submodule $(-f(m),m)\to (-f(m)-g(m),m)=(-f'(m),m)$.(i.e. This induces automorphism of $X_{f}$ and $X_{f'}$.) Clearly this induces the same extensions of $B$ by $A$.
$\textbf{Q:}$ Why do I expect $X_f\cong X_{f'}$ as the difference of 2 maps $f,f'$ are up to homotopy?(It is not totally obvious if the book does not mention there is such an automorphism, though the book did say by twisting $P\to X$ map you can get automorphism. That is the reason why I lifted automorphism on $B\oplus P$ level rather than $X_f\to X_{f'}$ level.) I am asking for intuitive reason for why this is the case without appealing to automorphism of $P\oplus B$. In particular for general extensions of length $n$, this construction also gives the well-definedness of the map of length $n-$extensions to $Ext^n(A,B)$.