Let $\pi: E \to M$ be a smooth vector bundle of rank $n$, and suppose $s_1, \ldots, s_m$ are independent smooth local sections over an open subset $U \subset M$.
Can I prove the "basis extension theorem" for smooth sections? That is, for each $p \in U$, there are smooth sections $s_{m+1},\ldots,s_n$ defined on some neighborhood of $V$ of $p$ such that $(s_1, \ldots, s_n)$ is a smooth local frame for $E$ over $U \cap V$.
Yes. Let $p:E\to M$ denote the fibration. Let $W\to U$ be the subbundle of $p^{-1}(U)\to U$ given by $W=\coprod _{p\in U}span\{s_{1}(p),...,s_{m}(p)\}$. Since the secions are smooth, this does indeed define a subbundle. Now let me give you an idea of how to proceed. You want to look at the normal bundle. If you fix a metric $g$ on $M$, there is a natural choice of such a bundle. Namely you can define $N=\coprod_{p\in U}N_{p}$ where $N_{p}$ is the orthogonal complement of $W_{p}$. Check (using the metric) that this is indeed a vector bundle. Then $N$ must be trivial, since $U$ is contractible. Choose a global section of the frame bundle of $N$ and your done!. You can do this without fixing a metric on $M$ too, you just need to use the quotient bundle $p^{-1}(U)/W$.