A surjective homomorphism between finite free modules of the same rank

You can show that every commutative ring is stably finite (see Lam's Lectures on Modules and Rings first 10 pages or so) which means that if $R^n\cong R^n\oplus N$, then $N=0$.

If you have a surjection $f:M\rightarrow M'$, then $M/\ker(f)\cong M'$, but $M'$ being projective implies that $0\rightarrow \ker(f)\rightarrow M\rightarrow M/\ker{f}\rightarrow 0$ splits, and so $M\cong \ker(f)\oplus M'$, whence $\ker{f}=\{0\}$.