There is something I don't understand in the definition(s) of a vector field.
A vector field on $\mathbb{R}^n$ is a continuous map $X: \mathbb{R}^n \to \mathbb{R}^n$.
Considering an arbitrary smooth $n$-manifold $M$. A vector field on M is a continuous section of the tangent bundle $TM$, i.e. a continuous map $X: M \to TM$ such that $p \circ X = \mathrm{id}_M$ (where $p: TM \to M$ is the bundle projection).
Now, here is my question. Consider an arbitrary continuous map $f: \mathbb{R}^n \to \mathbb{R}^n$, by definition, $X \circ f$ is again a vector field on $\mathbb{R}^n$, right ? However, consider now that $M = \mathbb{R}^n$. If we consider again the map $f$, by definition, I cannot see why $X \circ f$ is now a vector field on $M$ (due to the fact that $p \circ (X \circ f) = \mathrm{id}_M$ is not true in general).
So, I see two options:
1- The notion of "vector field on a smooth manifold" does not aim to generalize the notion of "vector field on $\mathbb{R}^n$", and I find it really strange. As differential geometry generalizes what we can do in Euclidean spaces, it seems weird to accept that those definitions of "vector field" are just different. I would find it more natural to define a notion of "vector field on a smooth manifold" which stick directly to the definition of "vector field on $\mathbb{R}^n$", as a particular case.
2- We can "fix the problem" and I do not see clearly how. More generally, if we suppose that $TM$ is trivial (i.e. if $TM$ is isomorphic, as a vector bundle, to $M \times \mathbb{R}^n$), why could we say that a vector field on $M$ is a map from $M$ to $\mathbb{R}^n$ ? I mean, even if, after identification, any vector field $X$ on $M$ is such that $X(m) = (m,v)$, for some $v \in \mathbb{R}^n$, there is still a problem, no ?
Thanks for your help !
Maps $M\to \mathbb R^n$ are in bijection with sections $M\to M\times \mathbb R^n$ over $M$, this is how the two notions are translated into each other. Explicitly, the bijection is obtained by sending a map $X\colon M\to \mathbb R^n$ to $(\operatorname{id}, X)$, and it's converse is given by postcomposing a section with the with the projection onto $M$.