Let $M$ be a manifold in $\mathbb R^n$. Observe that if $\varphi : M \to M$ is a diffeomorphism, $v \in T_p M$ and $f$ is a differentiable function in a neighborhood of $\varphi(p)$, we have $$ (d\varphi(v)f)\varphi(p) = v(f \circ \varphi)(p). $$ Indeed, let $\alpha : (-\varepsilon,\varepsilon) \to M$ be a differentiable curve with $\alpha'(0)=v$, $\alpha(0)=p$. Then $$ (d\varphi(v)f)\varphi(p) = \frac d{dt}(f \circ \varphi \circ \alpha) \Bigg\vert_{t=0} = v(f \circ \varphi)(p). $$
(from do Carmo, Riemannian geometry, page 26)
I am trying to justify the last very last equality. When I applied the chain rule, I have that $$ \frac d{dt}(f \circ \varphi \circ \alpha) = f'(\varphi \circ \alpha(t))\varphi'(\alpha(t))\alpha'(t). $$ Substituting in $t=0$ gives $$ \frac d{dt}(f \circ \varphi \circ \alpha)\bigg\vert_{t=0} = f'(\varphi(p))\varphi'(p)v, $$ but this is not the desired final expression $v(f \circ \varphi)(p)$. I am getting prime symbols, but $v(f \circ \varphi)(p)$ does not.
I know the post is old but maybe someone might stumble across and find this useful.
This is a usual misunderstanding when talking about tangent vectors as operators.
Now in our setting we have the following equality that we both agree on: $$d\varphi(v)(f)(\varphi(p))=\frac{d}{dt}f\circ\varphi\circ \alpha \Big|_{t=0}$$
We have that $\alpha’(0)=v$, thus by the very definition of $v$ (as an operator acting on a function) we obtain: $$v(f)=\frac{d}{dt} f\circ \alpha \Big|_{t=0}$$ So by definition of the tangent vector applied to $f\circ \varphi$ we see:
$$v(f\circ \varphi)= \frac{d}{dt} (f\circ\varphi\circ \alpha)\Big|_{t=0}= d\varphi(v)(f)(\varphi(p))$$ In this case there is nothing to justify it is all a matter of definitions :)