I am reading do Carmo's Riemannian Geometry, and he defines the tangent bundle $TM$ of a differentiable manifold $M$ as $\{ (p, v) : p \in M, v \in T_pM\}.$ That's fine. Next, he defines (on page 25):
A vector field $X$ on a differentiable manifold $M$ is a correspondence that associates to each point $p \in M$ a vector $X(p)$ $\in T_pM$. In terms of mappings, $X$ is a mapping of $M$ into the tangent bundle $TM$. The field is called differentiable if the mapping $X: M \to TM$ is differentiable.
So on one hand, he is saying that $X(p) \in T_p M$, which means that $X(p)$ is some tangent vector $v$. However, he then says that $X$ maps into $TM$, which means that $X(p) = (p, v)$ for some $v \in T_p M$. Am I missing something, or is he contradicting himself?
Syntactically, if $X:M\to TM$ is a vector field and $p\in M$, then $X(p)=(p,v)$ for some $v\in T_p M$ because $TM$ is a disjoint union of all $T_p M$'s, i.e. formally an object like $\bigcup_p \{p\}\times T_p M$ (but if you write it like this you should specify the proper topology that is meant).
But remember that $X$ is a section, i.e. $\pi \circ X=\textrm{id}$. This means that $X$ always sends $p$ to some pair $(p,\cdot)$, so the first element of it is readily determined. So in some sense it is redundant to write $X(p)=(p,v)\in \{p\}\times T_p M$ and we just write $X(p)=v\in T_p M$, slightly abusing notation for brevity.
It's nice to see someone else got puzzled by this too (like I was in the beginning). Rest assured, it's easy to get over it, as it never really causes any problems.