Can faithful flatness of a sheaf of modules be checked on stalks?

$\DeclareMathOperator{\mod}{Mod} \def\F{\mathcal{F}} \def\O{\mathcal{O}} \def\M{\mathcal{M}}$Yes, the converse holds. It follows from the following technical Lemma, see Proposition below. Given a topologically space $X$ and points $x,y\in X$, we write $x\rightsquigarrow y$ and say $x$ is a specialization of $y$, or $y$ is a generalization of $x$, if every neighborhood of $y$ is also a neighborhood of $x$, i.e., $y\in\overline{\{x\}}$ [SP, 0061]. The preorder $(X,\rightsquigarrow)$ is the specialization preorder. Write $[x]$ for the equivalence class of points in $X$ that are topologically indistinguishable from $x$, i.e., the points $y\in X$ satisfying $x\rightsquigarrow y$ and $y\rightsquigarrow x$. Note that for any sheaf $\F$ on $X$ we have $\F_x=\F_y$ whenever $[x]=[y]$.

Lemma. Let $X$ be a ringed space. Let $x\in X$ and let $M^\bullet$ be a chain complex of $\O_{X,x}$-modules. Then there is a chain complex $\M^\bullet$ of $\mathcal{O}_X$-modules whose stalk at $y\in X$ equals $$ \M^\bullet_y=\begin{cases} M^\bullet&y\in [x],\\ 0&\text{otherwise}. \end{cases} $$ In particular, if $X$ is Kolmogorov, $$ \M^\bullet_y=\begin{cases} M^\bullet&y=x,\\ 0&\text{otherwise}. \end{cases} $$

Proof. Write $p_x:(\{x\},\mathcal{O}_{X,x})\to (X,\O_X)$. Taking stalk of $p_{x,*}M^\bullet$ gives $M^\bullet$ at each point of $\overline{\{x\}}$ and gives the zero complex outside of this set [SP, 009B]. Suppose there were an $\O_X$-module $\F$ such that $$ \F_x=\begin{cases} \O_{X,x}&y\in [x],\\ 0&\text{otherwise}. \end{cases} $$ Then $\M^\bullet=\F\otimes_{\O_X}p_{x,*}M^\bullet$ does the job by [SP, 01CB]. Let us describe one such $\F$. Given $y\in X$, write \begin{gather} U_y=X\setminus\overline{\{y\}},\\ i_y:U_y\hookrightarrow X,\\ j_y:\overline{\{y\}}\hookrightarrow X. \end{gather} Define $$ \label{eq}\tag{1} \F=j_{x,*}\O_{X,x} \underset{\O_X}{\otimes} \bigotimes_{\O_X}^{ \substack{x\rightsquigarrow y\\ y\not\rightsquigarrow x} } i_{y,!}\O_{U_y}, $$ where:

1. $j_{x,*}\O_{X,x}=j_{x,*}j_x^{-1}\O_X$ is a sheaf of $\O_X$-algebras via the counit of $j^{-1}_x\dashv j_{x,*}$.

2. $i_{y,!}\O_{U_y}$ is a sheaf of $\O_X$-modules in the obvious way (but not of $\O_X$-algebras, for $i_{y,!}\O_{U_y}$ is not even a sheaf of rings in general).

3. The (possibly infinite) tensor on the right of \eqref{eq} is over all $y\in X$ that are specializations but not generalizations of $x$ [SP, 0061].

4. The infinite tensor product of $\O_X$-modules is the filtered colimit over the tensor products of $\O_X$-modules of finitely many factors (see infinitary tensor product), plus sheafification.

Taking stalks at $z\in X$ gives: $$ \label{stalk}\tag{2} \F_z=(j_{x,*}\O_{X,x})_z \underset{\O_{X,z}}{\otimes} \bigotimes_{\O_{X,z}}^{ \substack{x\rightsquigarrow y\\ y\not\rightsquigarrow x} } (i_{y,!}\O_{U_y})_z $$ since taking stalks commutes with finite tensor products [SP, 01CB] and hence with infinite tensor products for talking stalks commutes with any colimit (taking stalks at $z$ is the module pullback functor along $(\{z\},\mathcal{O}_{X,z})\to (X,\O_X)$, which is a left adjoint). We have that $(j_{x,*}\O_{X,x})_z$ equals $\O_{X,x}$ if $z\in \overline{\{x\}}$ and vanishes otherwise [SP, 009B]. Thus it suffices to see that for $z\in\overline{\{x\}}$ the infinite tensor product in \eqref{stalk} equals $\O_{X,x}$ if $z\in [x]$ and vanishes otherwise.

On the one hand, if $z=x$ then $y\not\rightsquigarrow x$ entails $z\in U_y$, so $(i_{y,!}\O_{U_y})_z=\O_{X,x}$ [SP, 00A5] and thus the infinite tensor product in \eqref{stalk} equals $\O_{X,x}$ for it is a filtered colimit of a diagram constantly $\O_{X,x}$. On the other hand, for $z\in\overline{\{x\}}\setminus [x]$, we have $x\rightsquigarrow z$ (by definition) and $z\not\rightsquigarrow x$ ($z$ is topologically distinguishable from $x$ by hypothesis). Therefore $y=z$ occurs in the infinite tensor product in \eqref{stalk}, i.e., the factor $(i_{z,!}\O_{U_z})_z=0$ shows up, hence the infinite tensor product vanishes (in the filtered diagram of the colimit giving the infinite tensor there is a cofinal subdiagram where everything vanishes). $\square$

Proposition. Let $X$ be a ringed space. An $\O_X$-module $\F$ is faithfully flat if and only if $\F_x$ is a faithfully flat $\O_{X,x}$-module for every $x\in X$.

Proof. The 'iff' pertaining flatness holds [SP, 05NE]. We now see the 'iff' pertaining faithful flatness. The implication to the left is argued in the original question above. Assume then $\F$ is a flat $\O_X$-module and that there is $x\in X$ such that $\F_x$ is not a faithfully flat $\O_{X,x}$-module. Then there exists a complex of $\O_{X,x}$-modules $M^\bullet$ that is not acyclic but such that $M^\bullet\otimes_{\O_{X,x}}\F_x$ is acyclic. Using the Lemma, we get a complex of $\O_X$-modules $\M^\bullet$ whose stalk at $y\in X$ is $M^\bullet$ if $y\in [x]$ and vanishes otherwise. Then $\M^\bullet$ is not acyclic but $\M^\bullet\otimes_{\O_X}\F$ is acyclic. $\square$