Does the geometric version of Nakayama's lemma hold for smooth manifolds? Consider the following geometric formulation of Nakayama's lemma.
Proposition. Let $F$ be a quasi-coherent sheaf locally of finite type on a scheme $X$. Consider the quotient map $\pi:F_x\to F_x\otimes\Bbbk (x)$. Given $s_1,\dots ,s_n\in F_x$, suppose their image generates $F_x\otimes \Bbbk (x)$. Then the $s_i$ extend to a neighborhood $U\subset X$ of $x$ on which they define a surjective arrow $$(\mathcal O _X|_U)^n\overset{(s_1,\dots ,s_n)}{\longrightarrow}F|_U\to \bf 0$$ on $U$. When this holds, we say $s_1,\dots ,s_n$ generate $F$ over $U$.
Let $(M,\mathcal T_M)$ be a manifold with the sheaf of sections of its tangent bundle. The $x$-fiber of $\mathcal T_M$ is the vector space of tangents at $x$. The $x$-stalk is the module of germs at $x$ of vector fields. Does "Nakayama" hold for $(M,\mathcal T_M)$?
 A: Yes, this is true for any vector bundle $E$ on a smooth manifold $M$.  Let $s_1,\dots,s_n$ be sections of $E$ in a neighborhood of $x$ whose values at $x$ span the fiber $E_x$.  We may assume that the values $s_1(x),\dots,s_n(x)$ are also linearly independent (if not, take a maximal linearly independent subset), so $n$ is the rank of $E$.  Now choosing a local trivialization of $E$, we can think of $(s_1,\dots,s_n)$ as a map $A:U\to\mathbb{R}^{n\times n}$ for some open neighborhood $U$ of $x$.  Since $s_1,\dots,s_n$ are linearly independent at $x$, $A(x)$ is an invertible matrix.  The set of invertible matrices is open in $\mathbb{R}^{n\times n}$, so $A(y)$ is invertible for all $y$ in some neighborhood $V$ of $x$.
Now let $s$ be any section of $E$ over $V$, which we think of as a map $V\to\mathbb{R}^n$ via our local trivialization.  For all $y\in V$, $s(y)=A(y)A(y)^{-1}s(y)$.  The columns of $A(y)$ are just $s_1(y),\dots,s_n(y)$, so this writes $s(y)$ as a linear combination $\sum f_i(y)s_i(y)$.  Here $f_i(y)$ is the $i$th coordinate of the vector $A(y)^{-1}s(y)$, which is a smooth function of $y$ (here we use that matrix inversion is smooth).  So, $s=\sum f_i s_i$ is a linear combination of the $s_i$ with coefficients in $C^\infty(V)$.  That is, the $s_i$ generate the sections of $E$ over $V$ as a module over $C^\infty(V)$.
