I have seen a number of proofs that the Lie bracket $[X,Y]$ of two vector fields $X$ and $Y$ is a vector field but I fail to see this from the definition of a vector as the directional derivative $v_\gamma$ along the curve $\gamma :\mathbb{R}\rightarrow M$.
In particular, given $\delta,\sigma$ two curves through a point $p\in M$ corresponding to the vector fields $X, Y$ respectively (not only at $p$, possibly along an open interval of the curve parameters including $p$), can we construct a curve $\gamma$ such that at the point $p$ it corresponds to the Lie bracket $[X,Y]$?
If not, then what would be a more demonstrative proof using the definition of vector fields as smooth sections of the tangent bundle?
Edit: I think I should clarify why I'm struggling with it and where I need it. Sometimes I see definitions of certain vector(tensor) fields are given as they act on some arguments. For example, in this particular case, $[X,Y]: C^\infty(M)\rightarrow C^\infty(M)$ is an $\mathbb R$-linear map such that $[X,Y]f:=X(Yf)-Y(Xf)$. However, I also understand that the set all $\mathbb R$-linear maps $C^\infty(M)\rightarrow C^\infty(M)$ is not the same as the set of all (iii)smooth (ii)sections of the (i)tangent bundle. My effort here is to show property (i) for the Lie bracket. Properties (ii) and (iii) are clear to me.
A similar case that I understand is when proving that set of all such sections is an $\mathbb R$-vector space, I had to prove that the pointwise sum two vector fields $X$ and $Y$ is a vector field. It (i) pointwise belongs to $T_pM$ because sum of two vectors at $p$ is a vector at $p$, is (ii) a section: trivial and (iii) smooth because sum of two smooth functions is smooth under the corresponding topologies and charts.
Best Answer
There are usually two definitions for tangent vectors: directional derivatives along curves and derivations of $C^\infty(M)$. While it is somewhat annoying to show that both definitions are equivalent, the fact remains that for some problems one definition is easier to work with than the other (for example, to show that $\mathfrak{X}(M)$ is a Lie algebra, the derivation definition is much easier to work with, but for finding the derivative of the multiplication map in a Lie group the curve definition is much easier to work with), so it is generally useful to be familiar with both definitions and establish early on that they are in fact equivalent.
Now, if we assume the derivation definition, your whole question just turns into a calculation: let $X,Y\in\mathfrak{X}(M)$, and let $f,g\in C^\infty(M)$, then $$ X(Y(fg)) - Y(X(fg)) = X(Y(f)g) + X(fY(g)) - Y(X(f)g) - Y(fX(g))$$ $$ = X(Y(f))g + Y(f)X(g) + X(f)Y(g) + fX(Y(g)) - Y(X(f))g - X(f)Y(g) - Y(f)X(g) - fY(X(g))$$ $$ = X(Y(f))g - Y(X(f))g + fX(Y(g)) - fY(X(g))$$ $$ = [X,Y](f)g + f[X,Y](g),$$ so that (+the $\mathbb{R}$-linearity properties) $[X,Y]$ is a derivation $C^\infty(M)\to C^\infty(M)$, which makes it into a vector field.