$\newcommand{Ext}{\operatorname{Ext}}\newcommand{Hom}{\operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r \in R$.
Applying snake lemma to the following diagram:
$$\begin{array}{c} 0 & \longrightarrow & A~~ & \longrightarrow & B~~ & \longrightarrow & C~~ & \longrightarrow & 0 \\ && \downarrow r && \downarrow r && \downarrow r \\ 0 & \longrightarrow & A~~ & \longrightarrow & B~~ & \longrightarrow & C~~ & \longrightarrow & 0 \end{array}$$
We obtain the exact sequence $0 \to A[r] \to B[r] \to C[r] \to A/rA \to B/rB \to C/rC \to 0$.
Now, assuming that $r$ is not a zero-divisor, $0 \to R \xrightarrow r R (\to R/rR \to 0)$ is a projective resolution of $R/rR$, and applying $\Hom(-,M)$ for $M \in \textbf{$R$-Mod}$ gives $0 \to \Hom(R,M) \xrightarrow{f \mapsto (x \mapsto f(rx))} \Hom(R,M) \to 0$, i.e. $0 \to M \xrightarrow r M \to 0$, so $\Ext^1_R(R/rR, M) = M/rM$ and $\Ext^n_R(R/rR, M) = 0$ for $n>1$, matching the exact sequence above.
If $r$ is a zero-divisor though, then according to this answer we do not get $M/rM$, but we need to apply $\operatorname{Hom}_R(-,M)$ to the exact sequence $0 \to rR \to R \to R/rR \to 0$ to get
$$0 \to M[r] \to M \to \Hom(rR,M) \to \Ext^1_R(R/rR,M) \to 0 \to \cdots$$
Now $\Hom(rR,M) \cong \{ m \in M \cdot \forall c \in R[r], c \cdot m = 0\}$ which I will notate as $M[R[r]]$.
So the exact sequence tells us $\Ext^1_R(R/rR,M) \cong M[R[r]]/rM$ instead of $M/rM$.
The exact sequence also tells us $\Ext^{n+1}_R(R/rR,M) \cong \Ext^n_R(rR,M)$ which may or may not be useful in computing $\Ext^\bullet_R(R/rR,M)$.
My question is: what is the role of the exact sequence $0 \to A[r] \to B[r] \to C[r] \to A/rA \to B/rB \to C/rC \to 0$ in the grand scheme of things involving $\Ext^\bullet_R(R/rR,-)$? If possible, I would also like to generalize to non-commutative ring with unity $R$ with $r \in Z(R)$.