Corollary 5.39, Lee - Introduction to Smooth Manifolds I am struggling with understanding how to prove Corollary 5.39 in Lee - Introduction to Smooth Manifolds.

I found an answer on this post: A characterization of tangent space to level set of a smooth submersion, but I do not understand the computation done there. Why do we have $d\Phi_p^i(v) = v\Phi^i$? By definition of the pushforward, for $f \in C^\infty(\mathbb{R})$ we have $d\Phi_p^i(v)f = v(f\circ\Phi^i).$ But by the linked answer, it seems we would also have $d\Phi_p^i(v)f = (v\Phi^i)f$. But $(v\Phi^i)f \in C^\infty(\mathbb{R})$, not $\mathbb{R}$. What am I missing here?
 A: We know that $d\Phi_p^i(v)$ is a vector in $T_{\Phi^i(p)}\mathbb{R} = \text{span }\Big(\frac{d}{dt}\big|_{\Phi^i(p)}\Big)$. So $d\Phi^i_p(v) = a \frac{d}{dt}\big|_{\Phi^i(p)}$ for some real number $a$. Since we often identify one dimensional vector $\lambda \, \frac{d}{dt}\big|_{t}$ with its component $\lambda$ itself, so (in our case) we write $d\Phi^i_p(v) = a$. But $a$ is exactly $v\Phi^i$ because
$$
a = \Big( a \, \frac{d}{dt}\Big|_{\Phi^i(p)} \Big) \text{Id}_{\mathbb{R}} = d\Phi^i_p(v) \,\text{Id}_{\mathbb{R}} = v(\text{Id}_{\mathbb{R}} \circ \Phi^i ) = \color{blue}{v \Phi^i}.
$$
So we can write $d\Phi^i_p(v) = v \Phi^i$.

The claim in the corollary still holds whether or not we apply this identification. Personally, i prefer to prove the corollary (without identification) as : $d\Phi_p(v) = 0 \in T_{\Phi(p)}\mathbb{R}^k$ if and only if
$$
0=d\Phi_p(v)y^i =v (y^i \circ \Phi) = v\Phi^i,\quad i=1,\dots,k, 
$$
where $y^i : \mathbb{R}^k \to \mathbb{R}$ are standard coordinate functions of $\mathbb{R}^k$.

