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.