Let $(X,\mathcal O_X)$ be a ringed space and $M$ an $\mathcal O _X$-module. The stacks project defines what it means for a family of global sections of $M$ to generate it by asking the associated arrow below to be epic. $$\coprod \nolimits _{i \in I} \mathcal{O}_ X \longrightarrow M$$
In the following lemma it is stated that a family of global sections generates $M$ iff for all $x\in X$ its germs generate the stalk containing them.
Now I am confused about two geometric formulations of Nakayama. In this MO answer it is stated that Nakayama implies a basis for the fiber lifts to a generating family for the corresponding stalk. On the other hand, e.g 4.B in this note by Vakil states that Nakayama implies a spanning set for a fiber lifts to a family of local sections over some open $V\subset X$ which generate $M|_V$.
Isn't the latter strictly stronger than the former? For instance, if we have a generating set for $M|_V$, then aforementioned lemma seems to imply we have a generating set for each stalk in $V\subset X$, not just the stalk associated with the fiber.
What am I missing here?
Perhaps a proof of 4B will be illuminating. 4B starts from the result of the MO answer and then strengthens it (but of course they're still equivalent). Well actually 4B is in fact strictly stronger in the sense that instead of assuming a locally noetherian scheme and a coherent module, it only assumes we start with a finite type module (finitely generated on affines) over an arbitrary scheme, which is enough to apply Nakayama's lemma, since all we need is that the stalk of the module is finitely generated over the local ring at a point.
In 4B we are given a finite type sheaf $\newcommand\calF{\mathcal{F}}\calF$ on a scheme $X$, a point $x\in X$, and a neighborhood $U$ of $x$ with sections $a_1,\ldots,a_n\in\calF(U)$ such that their images generate $\newcommand\tens\otimes \calF\tens k(x)$. Then Nakayama's lemma implies that their images generate the stalk of $\calF$ at $x$ as well. Choose $U'$ an affine open subset of $U$, and let $M=\calF(U')$. Identify $a_1,\ldots,a_n$ with their images in $U'$, and extend this set with elements $b_1,\ldots,b_m$ to a generating set for $M$ using that $M$ is f.g. since $\calF$ is finite type. Let $N$ denote the submodule of $M$ generated by the $a_i$, and consider $M/N$. Since $(M/N)_x=M_x/N_x=0$, for each $b_i$ there exists $\alpha_i\in \newcommand\calO{\mathcal{O}}\calO(U')$ such that $\alpha_i\not\in\newcommand\mm{\mathfrak{m}}\mm_x$, and $\alpha_ib_i\in N$. Letting $\alpha = \prod_{i=1}^m \alpha_i$, we have that $(M/N)_\alpha = M_\alpha/N_\alpha = 0$, so $U'_\alpha$ is a neighborhood of $x$ where the images of the $a_i$ generate $\calF(U'_\alpha)$.
This shouldn't be terribly surprising though. Try thinking of the module as analagous to a vector field on a smooth manifold. Then if we have a family of sections on an open set that span the fiber at some point, there should be some neighborhood of that point where the sections of the vector bundle generate all of the fibers of the points in that neighborhood, since we can reduce to a local trivialization of the bundle near the point, so that we can pretend everything is in $\Bbb{R}^n$. Then we take the determinant of the sections. The determinant is a polynomial, so the set of points where the determinant of the sections is 0 (i.e. the set of points where the sections don't span the fiber) is a submanifold of codimension 1 (also it's closed). Thus if the sections generate the fiber of the vector bundle at a point, then they generate the fibers of the points in some open neighborhood of that point. Perhaps it will be more intuitive thinking of it this way? I'm not sure. Hope this helps.
In fact I'm pretty sure one can more or less directly generalize this to schemes by replacing the determinant with the $n$th exterior power of the dual sheaf or something similar? I'd have to work that one out more carefully.