Here is the definition i found in my book on topology and geometry:
"A vector field $\zeta$ on an n-dimensional manifold $M$ is a function such that $\zeta(p)\in T_pM$ and such that given local coordinates $x_1, ... ,x_n$ near p such that $\zeta(p) = \sum_i a_i(p)\dfrac{\partial}{\partial x_i}$ the $a_i$ are smooth functions."
All of this seems a little complicated and informal, why not say that a vector field on an n-dimensional manifold $M$ is a smooth map $M \rightarrow \mathbb{R}^{n}$ since you often regard $T_pM$ as $\mathbb{R}^{n}$ anyways? Or am i mistaken in that $T_pM$ is isomorphic to $\mathbb{R}^{n}$ for all such M?
Cheers.
Here's an example demonstrating the difference. Let $M=S^2$. Each tangent space $T_pM$ is isomorphic to $\Bbb R^2$, and the constant function $f:p\mapsto(1,0)$ is a smooth, nowhere-vanishing map $M\to\Bbb R^2$. But by the hairy ball theorem, every vector field on the sphere vanishes somewhere.
The problem is that the global topological structure of $S^2$ prevents us from identifying every point's tangent space with $\Bbb R^2$ in a continuous way. In other words, the tangent bundle $TM$ is not homeomorphic to the product space $M\times\Bbb R^2$.