Let $\mathcal A, \mathcal B$ be abelian categories such that $\mathcal A$ have enough injectives and projectives. Let $F: \mathcal A \to \mathcal B$ be an additive left exact functor, and let $R^i F$ be its right derived functors. Now we know that given a short exact sequence $0\to A \to B \to C \to 0$, if $F$ is covariant (resp. Contravariant) one has a long exact sequence $0\to F(A) \to F(B) \to F(C) \to R^1F(A) \to R^1 F(B) \to R^1 F(C) \to ... \to R^iF(A) \to R^iF(B) \to R^i F(C) \to R^{i+1} F(A) \to ...$ (resp. $0\to F(C) \to F(B) \to F(A) \to R^1F(C) \to R^1 F(B) \to R^1 F(A) \to ... \to R^iF(C) \to R^iF(B) \to R^i F(A) \to R^{i+1} F(C) \to ...$ )
Now my question is: Can we do this more generally when instead of a short exact sequence , I start with a finite projective resolution of on object in $\mathcal A$ ? In particular, let $\mathcal A=\mathcal B = R$-Mod, where $R$ is a commutative Noetherian ring. Let $M$ be a $R$-module and $0\to X_n \to ... \to X_1 \to X_0 \to M \to 0$ be a finite free resolution of $M$ (i.e. all the $X_j$ are free $R$-module) . Let $F(-)= Hom(N,-)$ for some $R$-module $N$. Then do we have a long exact sequence $0\to F(X_n) \to ... \to F(X_1) \to F( X_0 ) \to F(M ) \to R^1F( X_n) \to ... \to R^1F(X_1) \to R^1F(X_0 ) \to R^1F(M) \to ... \to R^i F(X_n ) \to ... \to R^iF(X_1) \to R^iF(X_0 ) \to R^i F( M) \to R^{i+1} F(X_n) \to ... $ ? If this is indeed true, then can we generalise this to only assume all the $X_j$ s are free except possibly $X_n$ ? Can we generalize the situation further where we know nothing about $X_j$ s ?
No (at least if you want the maps $R^iF(X_j)\to R^iF(X_{j-1})$ and $R^iF(X_0)\to R^iF(M)$ in the long exact sequence to be the natural ones induced by the maps in the original resolution).
In fact, even $F(X_1)\to F(X_0)\to F(M)$ need not be exact at $F(X_0)$. If it were, for $F(-)=\text{Hom}(N,-)$, then every map from $N$ to the kernel of $X_0\to M$ must factor through $X_1$. But if $N$ is equal to this kernel then this is not true unless the map from $X_1$ to the kernel is a split epimorphism.