Given a ring $R$, we can consider the following functors:
any $A\in Mod-R$ and choice of projective resolutions $P_\bullet(B)$ for every $B\in R-Mod$ defines a functor $Tor_n^R(A,-):R-Mod\to Ab$,
any $B\in R-Mod$ and choice of projective resolutions $Q_\bullet(A)$ for every $A\in Mod-R$ defines a functor $tor_n^R(-,B):Mod-R\to Ab$.
It is known that $Tor_n^R(A,B)\simeq tor_n^R(A,B)$ as abelian groups for every $A\in Mod-R$, $B\in R-Mod$.
Question 1: can we define a bifunctor $Mod-R\times R-Mod \to Ab$ that, such that $(A,B)\mapsto Tor_n^R(A,B)$?
This first question is related to the second comment darij grinberg made in this question at MO.
Question 2: If $R$ is a commutative ring, then given an $R$-module $A$, we can consider $Tor_n^R(A,-), tor_n^R(-,A): R-Mod\to Ab$. Are these functors naturally isomorphic?
I think the answer to this question is yes, provided that the choices $Q_\bullet$ and $P_\bullet$ are the same. But what if they are different? If they are not necessarily naturally isomorphic, is there any condition that guarantees this is the case?
Now for a bonus question. The Tor functor involves an arbitrary choice in its definition that makes me quite uneasy. Formally, it doesn't seem to be well defined, since if we write $Tor_n^R(A,-):R-Mod\to Ab$, we aren't really taking into account the choice of projective resolutions.
Question 3 (bonus): Is there a way to fix this? Has this been considered in some treatise on the Tor functor?
EDIT: I've striked down the third question since I think I now understand how it goes: $Tor_n^R(A,-)$ (or any derived functor for that matter) is defined as a functor such that this and that. You then prove that every pair of functors that satisfy those conditions are naturally isomorphic, so you can be at rest that even if your choice was a bit arbitrary, you're not losing much without considering the other choices of projective resolutions.