Differential of a function between manifolds The book we are using in class is Frank Warner Foundation of Differential Manifold and Lie Group. Let $M,N$ be two smooth $d$-dim manifold, the differential of a $C^\infty$ function $\phi:M\rightarrow N$ is defined by 
$$d\phi: M_m \rightarrow N_{\phi(m)}$$
For $v\in M_n$, and $g:N \rightarrow \mathbb{R}$ a smooth function, we define
$$d\phi(v)(g) = v(g\circ \phi).$$
Now for a smooth function $f:M \rightarrow \mathbb{R}$, I don't quite see why the book states
$$df(v)(g) = v(f) \frac{\partial}{\partial r}\bigg|_{r_0} g$$
from the definition above.  
Edit: Reading the answer here seems that if we plug $g(r) = r$, we will get the identity $df(v) =  v(f)$, but how do we know it will hold for all $g$?
 A: Just to elaborate on Xiao's answer; considering the fact that we refer to tangent vectors derivations then the differential or push forward map;
$$[f_{*,p}(X_p)] g = X_p( g \circ f); \ X_p \in T_pM, g \in C_{f(p)}^{\infty}(N), f: M \to N$$
here $X_p( g \circ f) \in T_{f(p)}N$ (why?). Well if you take $g,h \in C_{f(p)}^{\infty}(N)$ then;
$$[f_{*,p}(X_p)] (gh) = X_p(gh \circ f) = X_p(g \circ f \cdot h \circ f)$$
and now since $X_p$ is a derivation;
$$X_p(g \circ f \cdot h \circ f) = X_p(g \circ f) \cdot h(f(p)) + g(f(p)) \cdot X_p(h \circ f) $$
$$ \hspace{1.2in}= [f_{*,p}(X_p)]g \cdot h(f(p)) + g(f(p)) [f_{*,p}(X_p)] h$$
The linearity piece is also clear since $X_p$ is linear. Therefore; if you take $f: M \to \mathbb{R}$ and $(U,x^1,...,x^d)$ to be a chart about $p$ then;
$$\left\{\frac{\partial}{\partial x^1}\Bigr|_p,...,\frac{\partial}{\partial x^d}\Bigr|_p\right\}$$
is a basis for $T_pM$. Similarly, we can use the coordinate $t$ to parametrize a neighborhood of $f(p) \in \mathbb{R}$ and so $T_{f(p)}\mathbb{R}$ has basis vector;
$$\frac{\partial}{\partial t}\Bigr|_{f(p)} := \frac{d}{dt}\Bigr|_{f(p)}$$
Since $f_{*,p}$ is linear, it maps tangent vectors to tangent vectors i.e;
$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) = \alpha \frac{d}{dt}\Bigr|_{f(p)}$$
If we evaluate both sides at $t$ and use the definition of the differential we have;
$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) t = \alpha \frac{d}{dt}\Bigr|_{f(p)} t \Rightarrow \frac{\partial}{\partial x^i}\Bigr|_p (t \circ f) = \frac{\partial}{\partial x^i}\Bigr|_p f = \alpha$$
The above follows from the fact that the coordinate function $t$ picks out the first coordinate of the map $f$, which is real-valued, so that if just $f$. It now follows that;
$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) =\frac{\partial}{\partial x^i}\Bigr|_p f \frac{d}{dt}\Bigr|_{f(p)}$$
A: By definition $df(v)$ is a tangent vector in $\mathbb{R}_{r_0}$, since $\bigg\{\frac{\partial}{\partial r}\bigg|_{r_0}\bigg\}$
is a basis of the tangent space $\mathbb{R}_{r_0}$, then it can be written as
$$df(v) (\cdot) =  K\frac{\partial}{\partial r}\bigg|_{r_0}(\cdot)\quad \text{ for some } K\in \mathbb{R}.$$
To solve for this $K$, we can plug  the identiy function into the tangent vector, that will tell us $K = v(f)$.
