Let $\pi:E\to M$ be a (smooth) vector bundle.
10.12. Let $C$ be a closed subset of $M$ and $s:C\to E$ a (smooth) section of $\pi_C:E|_C\to C$. For each open subset $U \subseteq M$ containing $C$, there exists a (smooth) section $\bar{s}$ of $U$ such that $\bar{s}|_C=s$ and $\operatorname{supp}\bar{s}\subseteq U$.
10.15. If $C\subseteq M$ is a closed subset and $\{s_1,\dots,s_k\}$ are linearly independent (smooth) sections of $\pi_C:E|_C\to C$, there exists a (smooth) reference frame $\{\bar{s}_1,\dots,\bar{s}_k\}$ of some open subset $U$ such that $C\subseteq U$ and $\bar{s}_i|_C=s_i$ for all $i\in\{1,\dots,k\}$.
I've been able to prove the first, but I'm stuck in the other.
The statement was false as I initially formulated it. I edited the question. The proof is:
Thanks to 10.12, for any open subset $V\subseteq M$ such that $C\subseteq V$, we can find a set $\{\bar{s}_1,\dots,\bar{s}_k\}$ of sections of $V$ such that $\bar{s}_i|_C=s_i$ for all $i\in\{1,\dots,k\}$. Since this sections restricted to $C$ are linearly independent, they are linearly independent in some open subset $U\subseteq V$ such that $C\subseteq U$ because being linearly independent is an open property. This is because if the sections are linearly independent at a point, its determinant (as a function) isn't zero at that point, so it isn't zero in some open neighbourhood of the point.