Diffeomorphism $\phi : M \to M$. Why can it be written like this? Let $\phi : M \to M$ be a diffeomorphism from $M$ to $M$. Let $v \in T_pM$ and $f$ a differentiable function near $\phi(p)$. Then we have
$$ \Big( d\phi (v) (f) \Big)\phi(p) = v(f \circ \phi)(p) $$
Why is this true?
The proof is given in the book (Carmo page 26) as following
$$  \Big( d\phi (v) (f) \Big)\phi(p) = \frac{d}{dt}(f \circ \phi \circ \alpha)\Big|_{t=0} = v(f \circ \phi)(p) $$
where of course $\alpha : I \to M$ is a differentiable curve with $\alpha(0)=p$ and $\alpha'(0)=v$. I do not understand how this is true. I only know that $d\phi = \beta'(0) = \phi(\alpha'(0))$.
I am very confused on the way he writes the above stuff and I would appreciate some clarifications.
 A: Expanding my comment, as I have found some online text, by who I assume to be the same author. Using (most of) your notation, his definition of the tangent map was $$ (d\phi)_p(v)=\left.\frac{d}{dt}(\phi\circ\alpha)\right|_{t=0}. $$ This is our definition. Heuristically this says: "The pushforward of a tangent vector at $p$ is the tangent vector at $\phi(p)$ you can get by composing the curve your original tangent vector is tangent to with the map $\phi$ and taking that curve's tangent vector."
We know that if $\gamma$ is a curve to which $u$ is tangent at $\gamma(0)$, then $u[f]=d/dt(f\circ\gamma)|_{t=0}$.
In our case we know from the definition that $(d\phi)_p(v)$ is tangent to $\phi\circ\alpha$, so $$ (d\phi)_p(v)[f]=\left.\frac{d}{dt}(f\circ\phi\circ\alpha)\right|_{t=0}. $$
But rather than interpreting the right hand side as $f$ composed with $\phi\circ\alpha$, you can interpret it as $f\circ\phi$ (which is a scalar field) composed with $\alpha$. So, by the definition of tangent vectors, $$ \left.\frac{d}{dt}(f\circ\phi\circ\alpha)\right|_{t=0}=\alpha'(0)[f\circ\phi]=v[f\circ\phi]. $$
