R is a local ring with maximal ideal $\mathfrak{m}$ and residue field k. M is a finitely generated R-module with a projective cover: $0 \to$ N $\to$ F $\to$ M $\to 0$.
Tensor $0\to$ $\mathfrak{m}$ $\to$ R $\to $ k $\to 0$ by N; right exactness gives a natural isomophism $\tau_N$: N $\otimes_R$ k $\to$ N/$\mathfrak{m}$N
I am not sure how to reach the last natural isomorphism by right exactness. I don't think the answer here Showing that if $R$ is a commutative ring and $M$ an $R$-module, then $M \otimes_R (R/\mathfrak m) \cong M / \mathfrak m M$. is correct as M might not be flat.
Any help would be appreciated!
By tensoring your exact sequence by $N$ you get an exact sequence $$N\otimes_R\mathfrak{m}\to N\otimes_RR\to N\otimes(R/\mathfrak{m})\to0$$ by right-exactness. This means that $N\otimes(R/\mathfrak{m})$ is isomorphic to the cokernel of $\phi:N\otimes_R\mathfrak{m}\to N\otimes_RR$. The image of $\phi$ is generated by the elements of the form $m\otimes a$ with $n\in N$ and $a\in\mathfrak{m}$.
Now $N\otimes_RR$ is isomorphic to $N$ via the map taking $n\otimes a$ to $an$. Under this isomorphism the image of $\phi$ corresponds to the submodule $N'$ of $N$ generated by the $an$ for $a\in\mathfrak{m}$ and $n\in N$, that is $N'=\mathfrak{m}N$. So the cokernel of $\phi$ is isomorphic to $N/N'$ which is isomorphic to $N/\mathfrak{m}N$. But this cokernel is also isomorphic to $N\otimes_R(R/\mathfrak{m})\cong N\otimes_Rk$.