In p. 176 of John M. Lee's Introduction to smooth manifolds, 2nd edition, Lee defines vector fields along subsets $A \subseteq M$ of a smooth manifold with boundary $M$.
The part that intrigues me is the fact that he requires a smooth local extension at each point by vector fields, not just any smooth maps.
That makes you think there is a separate definition for smooth vector fields along arbitrary subsets.
I've been thinking about it and I think the requirement of local extension at each point by vector fields is unnecessary.
I'll write briefly about that here in the more general context of vector bundles.
I want to know (preferably from @Jack Lee) if I'm doing this correctly.
This is the definition that Lee suggest of smooth sections along arbitrary subsets.
Definition. Let $M$ be a smooth manifold with or without boundary, $\pi: E \to M$ a vector bundle of rank $k$, $A \subseteq M$ an arbitrary subset and $\sigma: A \to E$ a rough section, i.e. a map (not necessarily continuous) such that $\pi\circ\sigma = Id_A$.
We say $\sigma$ is smooth if $\sigma$ extends to a smooth local section of E in a neighborhood of each point.
See Definition of smoothness to have some context about the definition of smoothness along arbitrary subsets.
And this is what I'm thinking.
Theorem 1. Let $M$ be a smooth manifold with or without boundary, $A \subseteq M$ an arbitrary subset and $k \in \mathbb{N}$.
If $f: A \to \mathbb{R}^k$ is smooth then there exist an open neighborhood $U$ of $A$ and a smooth extension $\overline{f}: U \to \mathbb{R}^k$ of $f$.
Proof. Use partitions of unity.
Theorem 2. Let $M$ be a smooth manifold with or without boundary, $\pi: E \to M$ a vector bundle of rank $k$, $(\sigma_1, \dots, \sigma_k)$ a local frame for $E$ over the open subset $U \subseteq M$, $A \subseteq U$ an arbitrary subset and $\sigma: A \to E$ a rough section.
Write $\sigma = \tau^i\sigma_i$ for some component functions $\tau^i: A \to \mathbb{R}$.
Then $\sigma$ is smooth if and only if all $\tau^i$ are smooth.
Furthermore, if $\sigma$ is smooth then there exists a neighborhood $W$ of $A$ and a smooth extension $\overline{\sigma}: W \to E$ of $\sigma$ which is a section such that $W \subseteq U$.
Proof. The first part is just following the definitions (but being careful of using the right definition of smoothness for arbitrary subsets of smooth manifolds with boundary).
For the second part, assume $\tau = (\tau^1, \dots, \tau^k): A \to \mathbb{R}^k$ is smooth. By Theorem 1., there exists a neighborhood $W$ of $A$ and a smooth extension $\overline{\tau}: W \to \mathbb{R}^k$ of $\tau$ such that $W \subseteq U$.
Now simply define $\overline{\sigma} = \overline{\tau}^i\sigma_i$.
This is a smooth section and extends $\sigma$.
Theorem 3. Let $M$ be a smooth manifold with or without boundary, $\pi: E \to M$ a vector bundle of rank $k$, $A \subseteq M$ an arbitrary subset and $\sigma: A \to E$ a rough section.
If $\sigma$ is smooth then there exists an open neighborhood $U$ of $A$ and a smooth extension $\overline{\sigma}: U \to E$ of $\sigma$ which is a section.
Proof. By Theorem 2., we can pick for each $p \in A$, a neighborhood $U_p$ of $p$ and a smooth extension $\overline{\sigma}_p: U_p \to E$ of $\sigma|_{A \cap U_p}$ which is a section.
Now pick a smooth partitions of unity $(\psi_p)_{p \in A}$ subordinated to the open cover $(U_p)_{p \in A}$ of $U = \bigcup_{p \in A}U_p$.
Now define
\begin{equation*}
\overline{\sigma}(x) = \sum_{p \in A}\psi_p(x)\overline{\sigma}_p(x),
\end{equation*}
for $x \in U$.
This is a smooth section and extends $\sigma$.
Best Answer
As confirmed by John M. Lee, what I've wrote in the question is fine, though maybe a bit wordy. Therefore, I will mark this question as answered.