Tensor product of a free module commutes with arbitrary intersection

I was wondering if the following holds true:

Suppose that we have two modules $$M$$ and $$N$$ over a commutative ring $$R$$. Suppose that $$M$$ is a finitely generated free $$R$$ module. Suppose $$(N_i)_{i\in I}$$ are submodules of $$N$$, where $$I$$ is arbitrary. We identify each $$M \otimes N_i$$ with a submodule of $$M \otimes N$$, using that $$M$$ is flat and similarly identify $$M \otimes (\bigcap_{i \in I} N_i)$$ with a submodule of $$M \otimes N$$. Then $$\bigcap_{i \in I} (M \otimes N_i) = M \otimes (\bigcap_{i \in I} N_i)$$ as submodules of $$M\otimes N$$.

1) A proof of this was provided here in the case where $$M$$ is merely flat, but at the cost of only allowing $$I$$ to be finite.

2) In the answer for this question, it is shown that tensoring by $$M$$ commutes with limits when $$M$$ is even finitely generated and projective.

To be able to apply point 2) here, I was wondering whether I can regard the intersection of submodules $$N_i$$ as a limit of some functor $$\mathcal{F}$$ into $$R-\text{Mod}$$ (The notion of limit that I use is the one from the textbook of Aluffi in Chapter VIII).

I believe that this should be so, but my lack of background in category theory makes me feel uncertain about this claim. Of course, I understand why this is so if we were looking into the category of sets and $$N_i$$ were some subsets of a set $$N$$.

I believe that the intersection of two submodules of a modules can be regarded as a fibered product, and I believe fibered products are examples of limits. But I am not so sure about arbitrary intersections.

let $$M\cong R^n$$. Then $$M\otimes N\cong \oplus_{j=1}^nN$$ as $$R$$-modules. Let's see where the submodules u are looking for go. $$M\otimes N_i \cong \oplus_{j=1}^nN_i$$ under this identification. So $$\cap _iM\otimes N_i \cong \cap_i \oplus_{j=1}^nN_i=\oplus_{j=1}^n\cap_iNi\cong M\otimes \cap_iN_i$$. So in the identification the 2 submodules are equal. Thus the 2 submodules are indeed equal.