(Reproduced from there.)

Since ${R\choose k}$ is the coefficient of $x^k$ in the polynomial $(1+x)^R$ and ${M\choose n-k}$ is the coefficient of $x^{n-k}$ in the polynomial $(1+x)^M$, the sum $S(R,M,n)$ of their products collects all the contributions to the coefficient of $x^n$ in the polynomial $(1+x)^R\cdot(1+x)^M=(1+x)^{R+M}$.

This proves that $S(R,M,n)={R+M\choose n}$.

(Reproduced from there.)

Since ${R\choose k}$ is the coefficient of $x^k$ in the polynomial $(1+x)^R$ and ${M\choose n-k}$ is the coefficient of $x^{n-k}$ in the polynomial $(1+x)^M$, the sum $S(R,M,n)$ of their products collects all the contributions to the coefficient of $x^n$ in the polynomial $(1+x)^R\cdot(1+x)^M=(1+x)^{R+M}$.

This proves that $S(R,M,n)={R+M\choose n}$.