Let $R \subset S$ be an injective map of commutative rings, $M \subset N $ an inclusion of $R$-modules. We identify $M, N$ respectively as $M \otimes_R 1_S \subset N \otimes_R 1_S \subset N \otimes S$.
In this answer Haran elaborated that if $R \subset S$ a faithfully flat extension, then the intersection of $N \otimes_R 1_S $ and $ M \otimes_R S $ behaves as nice as one would naive expect, i.e., $$M \otimes_R 1_S = (N \otimes_R 1_S) \cap M \otimes_R S. $$
Remark: As Aphelli pointed out, the intersection inside $N \otimes_R S $ with $M \otimes_R S $ should be read as the intersection with the image of the induced map $M \otimes S \to N \otimes S $, which (if we not assume flatness for $\phi$ not need to be injective)
Question: What about a converse statement? Can this be used as sufficient criterion for faithfully flatness? (at least if we already assume $R \subset S$ to be flat?)
That means, if for all inclusions $M \subset N $ of $R$-modules, holds $M \otimes_R 1_S = (N \otimes_R 1_S) \cap M \otimes_R S $, then does it follow that $R \subset S$ is faithfully flat?
Assuming that $R\to S$ is injective and flat, the answer is yes:
Let $M\subset R$ be an ideal and let $N=R$. Then from the assumption we get that $M\otimes_R1_S=R\otimes_R1_S\cap M\otimes_RS\subset R\otimes_RS$. Via the obvious identifications, this becomes $M=R\cap MS\subset S$. Thus, if $MS=S$, then $M=R$. Since $M$ was arbitrary (and under the assumption that $S/R$ is flat to begin with), it follows that $S/R$ is faithfully flat.