Look in the bottom for the edited answer:
Is possible to prove that we could extend globally a local vector field $X$ defined on a open subset $U$ of a real paracompact manifold $M$? Is my congecture ok?
Let $U$ be the open subset of a smooth manifold $M$ and $X$ the vector field defined on $U$. Wlog there exist $V$ open subset containing $U$, so $\{V,\overline{U}^c\}$ is an open covering of $M$, and for a knowned theorem there exist a countable smooth partition of unity whose dominated the covering; $\{\psi_i\}_{i\in\mathbb{N}}$.
Let $J\subset\mathbb{N}$ such that $\sum_{i\in J}\psi_i(p)=0, \forall p\notin V$, so $\sum_{i\in J}\psi_i(p)=\sum_{i\in\mathbb{N}}\psi_i(p)=1, \forall p\in U$.
Our global smooth vector field is $X(p)\sum_{i\in J}\psi_i(p), p\in U$ and $\underline 0$ for $p\notin V$;
but I have a problem to extend $X$ in $V-U$ because $V$ may not be a domain of chart (in these case it will be trivial) but it could be the non-disjoint union of two domain of charts.
Apologize me if I'm writing wrong things
The answer to my first question is NO, see the comment. Now I'm asking when is possible to extend a vector field defined on a non-dense subset of a manifold in all the whole manifold?
Best Answer
Here is the correct statement: Suppose that $M$ is a smooth manifold, $U\subset M$ is an open subset and $X\in {\mathfrak X}(U)$ a (smooth) vector field. Then for every compact $K\subset U$ there exists a smooth vector field $Y$ on $M$ whose restriction to $K$ equals that of $X$. If this is what your notes use, I will write a proof. If not, think of the following example: $M={\mathbb R}^2$, $U=M -\{0\}$, $X=\frac{1}{x^2+y^2}\partial_x$. Then $X$ does not extend to $M$.