Showing that d$f(X)=X(f)$ Let $M$ be a manifold, $f:M\rightarrow\mathbb{R}$ a differentiable map and $X$ a differentiable vector field on $M$. I am struggling a bit understanding the equality
$$\text{d}f(X)=X(f)$$
I believe it is equivalent to
$$\text{d}_pf(v)=v(f)\quad\text{for every}\quad v\in T_pM$$
by making $v=X_p$ at each $p\in M$.
The rhs is a real number and the lhs is a vector in $T_p\mathbb{R}$ (which obviously can be identified with a real number but technically, is not a real number). But if we take the identity chart $t:\mathbb{R}\rightarrow\mathbb{R}$, doing some computations I get that
$$\text{d}_pf(v)=v(f)\left.\frac{\partial}{\partial t}\right|_{f(p)}\Rightarrow \text{d}f(v)=v(f)\frac{\partial}{\partial t}\Rightarrow \text{d}f(X)=X(f)\frac{\partial}{\partial t}$$
which makes more sense to me.
So my question is, is my reasoning correct and this first formula is just an abuse of notation or am I missing something?
 A: For any smooth function $f : M \to \mathbb{R}$ we define the 1-form $df$ by the rule $df(X) = X(f)$ for every $X \in TM$.  At any point $p$ this gives a real number $df_p(X) = X_p(f)$.  
On the other hand, for any differentiable map $F : M \to N$, where $N$ is another smooth manifold, we have the differential of $F$, also called the push forward, sometimes also denoted by $dF$;  it is a map of the tangent bundles $dF:TM \to TN$.  In this case $dF(X)$ is an element of $TN$ for each $X \in TM$.  (By definition, for any smooth function $\phi : N \to \mathbb{R}$ we set $dF(X)(\phi) = X(\phi \circ F)$, thus defining $dF(X) \in TN$ as a derivation.) 
Now, given $f: M \to \mathbb{R}$ we can view $f$ either as a smooth function, or if we like as a differentiable map between smooth manifolds.  In the former case $df$ is a one-form (it "eats" tangent vectors and spits out numbers).  In the latter case $df$ is a linear map $df : TM \to T\mathbb{R}$ (so that at each point $df$ "eats" tangent vectors of $M$ and spits out tangent vectors to $\mathbb{R}$).  The notation may be a little confusing, but typically context makes it clear which $df$ is being referred to.  
