How to use $5$-lemma to prove that $F(M) \otimes_RM' \overset{\simeq}{\longrightarrow} F(M \otimes_R M') $ is a (natural) isomorphism? I am describing the question details, though the main question is short as below.
Let $O$ be the ring of integers of the finite extension $K$ of the $p$-adic field $\mathbb{Q}_p$. Let $R$ be a finite $\mathbb{Z}_p$-algebra. Let $\bar K$ be the algebraic closure of $K$ and $G_K:=\text{Gal}(\bar K/K)$  Then consider a right exact functor $$F: \mathscr{C} \to \mathscr{D}$$ between the abelian categories $\mathscr{C}$ and $\mathscr{D}$, where $\mathscr{C}$ is the category of finite $O \otimes_{\mathbb{Z}_p}R$-modules and $\mathscr{D}$ is the category of continuous representations on finite $R$-module (I think any abelian categories is ok for the below question).
Question:  For a finite $O \otimes_{\mathbb{Z}_p}R$-module $M \in \mathscr{C}$ and any finite $R$-module $M'$, there is a natural isomorphism $$F(M) \otimes_RM' \overset{\simeq}{\longrightarrow} F(M \otimes_R M').$$
Some hints is given. It is asked to take a presentation of the module $M'$ and then to use the $5$-lemma to prove the above isomorphism. So I was trying the following way:
$$------------------------------------$$
My efforts:
For natural numbesr $m,n$, Take the finite presentation $R^{\otimes n} \to R^{\otimes m} \to M' \to 0$ of the finite $R$-module $M'$
and then do the following two things:
$(i)$ First take with tensor product $(O\otimes_{\mathbb{Z}_p} R)$-module $M$ and then apply the given right exact-functor $F$ to this finite presentation to obtain the exact sequence $$ F(M \otimes R^{\otimes n}) \to F(M \otimes R^{\otimes m}) \to F(M \otimes M') \to 0$$
$(ii)$ Take tensor product of the finite presentation $R^{\otimes n} \to R^{\otimes m} \to M' \to 0$ by $F(M)$ to obtain the exact sequence  $$F(M) \otimes_R R^{\otimes n} \to F(M) \otimes_R R^{\otimes m} \to F(M) \otimes_R M' \to 0.$$
I think now from $(i)$ and $(ii)$, we have the following commutative diagram to apply $5$-lemma:
\begin{align}
\matrix{
F(M \otimes_R R^{\otimes n}) &\to&F(M \otimes_R R^{\otimes m}) &\to&F(M \otimes_R M')&\to&0
\cr \downarrow f_1&&\downarrow f_2&&\downarrow {\color{red}{f_3}}&&\downarrow f_4&&
\cr F(M) \otimes_R R^{\otimes n}&\to&F(M) \otimes_R R^{\otimes m}&\to& F(M) \otimes_R M' &\to& 0}
\end{align}
I think using $5$-lemma, we need to show ${\color{red}{f_3}}$ is an isomorphism to answer our question. I think we need to define $f_1, f_2$ in order to satisfy $5$-lemma criteria. But here we have $4$ columns and we can use $4$-lemma at best instead of $5$-lemma. How do get the $5$-th column according to the given hints.
Am I doing correct so far?
Any guidance and help lease.
 A: Okay, I might as well answer. Whoever gave the hint "5-lemma" might have been using that term as a shorthand for "apply some standard homological algebra result", knowing that some application or other of the 5-lemma would get the job done. But that hint doesn't seem optimized.
Let $R$ be a commutative ring. For any category $\mathcal{C}$ enriched in the category of $R$-modules and with finite colimits, there's a sensible notion of tensoring an object $C \in \mathcal{C}$ by a finitely presented $R$-module $M$, and this is implicit in what you write: for a natural number $n$, define $C \otimes_R R^n$ to be $C^n$, the coproduct of $n$ copies of $C$, guided by the intuition
$$C \otimes_R R^n \cong C \otimes_R (R \oplus \ldots \oplus R) \cong (C \otimes_R R) \oplus \ldots \oplus (C \otimes_R R) \cong C \oplus \ldots \oplus C = C^n.$$
This definition is easily made functorial: if we have an $R$-module map $f: R^m \to R^n$, i.e., a matrix, then we get an induced action $C^m \to C^n$ of this matrix by utilizing the $R$-module enrichment. Then, if we have a finite presentation $R^n \to R^m \to M \to 0$ of an $R$-module $M$, we may define $C \otimes_R M$ by taking the cokernel of the induced map $C^n \to C^m$. It may be checked that this definition of $C \otimes_R M$ doesn't depend essentially on the chosen presentation. (Side comment is that if $R$ is Noetherian, then finite presentability is equivalent to being finitely generated.)
Now suppose given an right exact additive functor $F: \mathcal{C} \to \mathcal{D}$ between finitely cocomplete categories enriched in $R$-Mod. This means $F$ preserves finite direct sums and cokernels. In particular, for the $f_1$ in the diagram above, just transcribe the canonical isomorphism
$$F(C \otimes_R R^n) \cong F(C^n) \cong F(C)^n \cong F(C) \otimes_R R^n$$
where the middle isomorphism just results from the fact that an enriched functor must automatically preserve finite direct sums. Same for $f_2$.
Now apply $F$ to the cokernel of $C^n \to C^m$, i.e., to the exact sequence $C^n \to C^m \to C \otimes_R M \to 0$. That gives essentially your top horizontal sequence. But $F$, being right exact, preserves this exact sequence, so $F(C \otimes_R M)$ is isomorphic to the cokernel of the map $F(C^n) \to F(C^m)$, which we have isomorphically identified with the map $F(C)^n \to F(C)^m$. But the cokernel of the latter is $F(C) \otimes_R M$ by our definitions.
