Equality of two notions of tensor products over a commutative ring

Let $R$ be a commutative ring. Let $M$ and $N$ be $R$-modules. Let $M\otimes_R N$ be the tensor product in the former sense. Let $M\tilde\otimes_R N$ be the tensor product in the latter sense. We can regard $M\otimes_R N$ as an $R$-module in the obvious way. Then there exists the unique $R$-linear map $\psi\colon M\tilde\otimes_R N \rightarrow M\otimes_R N$ such that $\psi(m\tilde\otimes n) = m\otimes n$.

Let $Z$ be an abelian group. Let $f:M\times N \rightarrow Z$ be a bilinear map such that $f(rm, n) = f(m, rn)$.

There exists the unique group homomorphism $f^* \colon M \otimes_R N \to Z$ such that $f^*(m \otimes n) = f(m,n)$.

Then $f^*\psi\colon M\tilde\otimes_R N \rightarrow Z$ is a group homomorphism such that $f^*\psi(m \tilde\otimes n) = f(m,n)$. Since the set of $m\tilde\otimes n$ generates $M\tilde\otimes_R N$ as a group, The uniquness of $f^*\psi$ as a group homomorphism is clear.

[This answers a question raised in the comments above]

Let $R$ be a (not necessarily commutative) ring and let $M_R$ and ${}_RN$ be a right and a left $R$-module, respectively. Pick a free presentation $$F_1\to F_0\to M\to 0$$ of $M$ as a right $R$-module, that is, a short exact equence of the form above with $F_0$ andd $F_1$ free right $R$-modules and $R$-linear maps. This amounts to writing $M$ in terms of generators and relations.

Since the tensor product $(\mathord-)\otimes_RN$ is a right exact functor, applying it to the sequence above preserves exactacts, so we get a short exact sequence $$F_1\otimes_RN\to F_0\otimes_RN\to M\otimes_RN\to 0$$ This means, precisely, that $M\otimes_RN$ is the quotient of $F_0\otimes_RN$ by the (image of) $F_1\otimes_RN$.