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Let $M$ be a manifold and $TM=\coprod_{x\in M}T_xM$ be the tangent bundle. By definition, a vector field is an application \begin{align*} X: M&\longrightarrow TM\\ m&\longmapsto X_m\ni T_mM \end{align*} but I don't see concretely how it looks like. Let take for exemple $M=\mathcal S^2$, the unit sphere in $\mathbb R^3$. So if $X$ is a vector field of $M$, then $X_m$ is parallel to any curve embedding in $M$ passing through $m$ ? So, for exemple, if $X_{(0,0,1)}=(0,0,1)$, it's not in a vector field of $M$, is it true ?

Other thing, if $\dim M=n$, then $T_mM=\text{span}(\partial {x_1},...,\partial _{x_n})$. If we take for exemple $M=\mathbb R^2$, to me $T_0M=\mathbb R^2=span(e_1,e_2)$, so what would be $\partial _{x},\partial _y$ here since $T_0M=\text{Span}(\partial _x,\partial _y)$ ? Are there $e_1$ and $e_2$ ? if yes, how can $e_i$ can be a derivative ? Sorry for those confusion.

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    Related: Understanding tangent vectors. (Similar questions from a variety of angles have been asked many times over the years; perusing the "related" links may be helpful.) For the record, "yes, $X(0,0,1) = (0,0,1)$ is not tangent to the unit sphere", and a vector $v$ at a point $x$ defines a directional derivative operator, $$f \mapsto \frac{d}{dt}\bigg|{t=0} f(x + tv).$$Thus, $\partial{x} = e_{1}$ and $\partial_{y} = e_{2}$, just as you say. (Technically, $\partial_{x}$ is the constant vector field whose value is $e_{1}$ at each point.) – Andrew D. Hwang May 01 '16 at 13:55

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