This answer (as pointed out by Eric Wofsey in the comments) only works if we assume that $φ$ is natural in $N$, ie. that $φ ∘ (\mathrm{id} × f) = Ff ∘ φ$ for every $f : N → N'$. In that case the action of $F$ on morphisms coincides with what you get by the construction described below.
Yes, this is an important technical fact in category theory.
Given a functor $G : \mathscr D → \mathscr C$, if there is "locally" an object $FC ∈ \mathscr D$ and a universal morphism $η_C : C → GFC$ (ie. for every $f : C → GD$ there exists a unique $g : FC → D$ such that $Gg ∘ η = f$) for every $C ∈ \mathscr C$, then $F$ can be pieced together into a functor defined uniquely up to a unique isomorphism, and $F$ is left adjoint to $G$.
To see how this applies in your case, notice first that the universal bilinear map $φ_N : M × N → FN$ can be replaced by the universal linear map $ψ_N : N → \mathrm{Hom}_R(M, FN)$, $ψn := φ(-, n)$, so that we can take $\mathscr C = \mathrm{Mod}_R$, $\mathscr D = \mathrm{Mod}_R$, and $G = \mathrm{Hom}_R(M, -)$.
Here's the quick proof overview of the fact in the first paragraph.
First choose a universal morphism $η_C : C → GFC$ for every $C$. Now, for every $f : C → C'$ in $\mathscr C$ there is a unique morphism $g : FC → FC'$ such that $Ug ∘ η = η ∘ f$, and we can set $g = Ff$. Uniqueness will guarantee that $F$ constructed in this way is a functor, and $η : \mathrm{Id} ⇒ GF$ a natural transformation. Now given another choice $η'_C : C → GF'C$ of universal morphisms there will be a unique isomorphism $α_C : FC → FC'$ such that $αη = η'$, and this lets you prove that $α$ is natural. (Notice that $α∘Ff∘η = F'f∘α∘η$ = $η'$, so we have to have $α ∘ Ff = F'F∘α$ too.)
Technically, since you're already assuming your $F$ is a functor, you only need the last part of the proof, but the more general perspective is useful.
Edit: The answer assumes that $F$ is a functor $\mathrm{Mod}_R → \mathrm{Mod}_R$, since that's what you need to express the usual universal property of tensor product over a commutative $R$ (that $R$-bilinear maps $M × N → P$ correspond to $R$-linear maps $M ⊗_R N → P$), if you really want to consider $M ⊗_R N$ in $\mathrm{Ab}$, then you get a different universal property (that $R$-balanced maps $M × N → A$ correspond to additive maps $M ⊗_R N → A$), but it is again unique by the same argument (you can take $\mathrm{Hom}(M, -)$ for $G : \mathrm{Ab} → \mathrm{Mod}_R$, because $\mathrm{Hom}(M, A)$ has a canonical $R$-module structure).