In Differential Geometry, why does it make sense for a smooth vector field to act on a smooth function?

For instance consider the smooth vector field $$\phi: \mathscr{M} \longrightarrow T\mathscr{M}$$ that maps every point on a smooth manifold $\mathscr{M}$ to an element in the tangent space of $\mathscr{M}$.

A lot of books, after defining the idea of a smooth vector field, move on to talk about the space of all such vector fields $\Gamma(T\mathscr{M})$. It's further expressed that this space forms a $C^{\infty}(\mathscr{M})$ - module but in order to show this, they mysteriously talk about the tangent vector fields as functions from $C^{\infty}(\mathscr{M})$ to $C^{\infty}(\mathscr{M})$. Am I missing a step here? I thought the vector fields are functions on the manifold, i,e, the argument of the function is a point on the manifold. How can a vector field of this sort act on a smooth function? It doesn't sense to me. Can someone please clarify this for me. An elaborate answer would be much appreciated as I'm just a beginner in this subject.

• In differential geometry, a tangent vector $\vec{v} = |v|\hat{v}$ at a point $p$ is essentially an equivalent class of curves crossing that point $p$ along direction $\hat{v}$ with speed $|v|$. It act on any smooth functions defined over some neighbor of $p$ by taking directional derivatives along any member curve of corresponding equivalent class. In short, tangent vector $v$ is really directional derivatives in disguise. So it is natural to talk about action of a tangent vector field on a smooth function. – achille hui Nov 17 '16 at 12:19

Think about what a tangent vector $v$ at a point $x$ does: it takes in a function $f:M \to \mathbb{R}$ and computes the 'directional derivative' $v(f)$ in the direction of $v$ at the point $x$. In this sense, $v$ can be thought of as a function $v:C^\infty(M)\to \mathbb{R}$ (Which satisfies the product rule at $x$, and 'really' acts on smooth functions locally defined near $x$)
A map $V:M \to TM$, then, is a collection of these maps from $C^\infty(M)$ to $\mathbb{R}$; the natural question is then how these maps vary as we move around the manifold. That is, if $f\in C^\infty(M)$ is a function and $x,y\in M$, the values of the vector field $V_x$ and $V_y$ are tangent vectors at $x$ and $y$ respectively. Then naturally we can apply both of them to $f$; what should the relationship be between $V_x(f)$ and $V_y(f)$ be? A priori there's no reason to believe there really should be any relationship between them at all; but one way to define a smooth vector field (or, if you're using another definition of a smooth vector field, a reasonably straightforward theorem) is to say that the map $x\mapsto V_x (f)$, with $V$ and $f$ fixed and the point on $M$ allowed to vary, should be a smooth function. It's this function that we call $V(f)$, and this defines a map $V:C^\infty(M) \to C^\infty(M)$.
$\Gamma (TM)$ is a $C^{\infty}$-module simply because for every $f \in C^{\infty}(M)$ and $\phi \in \Gamma(TM)$, one can define a new vector field $f \phi$ defined as follows: for $x \in M$, $$(f \phi)_x = f(x) \cdot \phi_x.$$ Here the real number $f(x)$ is being multiplied to the tangent vector $\phi_x$. This defines an action of $C^{\infty}(M)$ on $\Gamma(TM)$.
On the other hand, the tangent space $T_xM$ can be defined as the set of all derivations on the germs of all smooth real-valued functions in a neighborhood of $x$. So, given $\phi \in \Gamma(TM)$ and $f \in C^{\infty}(M)$, we can define a smooth real-valued function $\phi(f)$ by $$\phi(f) (x) = \phi_x(f).$$ (Note that $\phi_x \in T_xM$ is a derivation on the set of germs of all smooth functions defined near $x$, of which $f$ is a member) Thus we have a map $$\phi : C^{\infty}(M) \to C^{\infty}(M) , f \mapsto \phi(f).$$
You are correct, smooth vector fields are smooth sections of tangent bundle. Concretely, $X\colon M\to TM$ must be smooth and to be a section, $X(m)$ must be some tangent vector at $m$, for any $m\in M$. Usually we denote that tangent vector with $X_m$. Now, tangent vectors act on $C^\infty(M)$ by definition. Thus, we can define map $\bar X\colon C^\infty(M)\to C^\infty(M)$ with $(\bar Xf)(m) = X_mf$. It turns out that $\bar X$ is linear operator and moreover derivation of algebra. One can show that converse is true as well, any derivation $D$ of algebra $C^\infty(M)$ gives rise to smooth vector field.