Lee's Introduction to Riemannian manifolds (2nd ed.) introduces the vertical tangent vector space as follows (Page 21):
Here, $\tilde{M}_p = \pi^{-1}(p)$ for $p\in M$.
Suppose $\tilde{M}$ and $M$ are smooth manifolds, $\pi:\tilde{M}\to M$ is a smooth submersion, ..., At each point $x\in \tilde{M}$, we define two subspaces of the tangent space $T_x \tilde{M}$ as follows: the vertical tangent space at $x$ is $\begin{equation} V_x = \textrm{Ker}(d\pi_x) = T_x(\tilde{M}_{\pi(x)})\tag*{}\end{equation}$ (that is, the tangent space to the fiber containing $x$). (The definition for the horizontal tangent space follows.)
The question is that I don't understand why $\textrm{Ker}(d\pi_x) = T_x(\tilde{M}_{\pi(x)})$ is true. Since the book does not give proof, the following is my attempt.
My attempt Suppose $x\in \tilde{M}$. Let's first prove $T_x(\tilde{M}_{\pi(x)})\subset \textrm{Ker}(d\pi_x)$. For any $v\in T_x(\tilde{M}_{\pi(x)})$, take a smooth curve $\gamma:[0,1]\to \tilde{M}_{\pi(x)}$ such that $\gamma(0)=x, \gamma'(0)=v$. Then, by definition of $M_{\pi(x)}=\pi^{-1}(\pi(x))$, we have $\pi\circ \gamma(t)=\pi(x)$ i.e. $\pi \circ \gamma$ is a constant curve on $M$. This implies that $(\pi\circ \gamma)'(0) = d\pi_{x}(\gamma'(0)) =d\pi_{x}(v)=0$, so $v\in \textrm{ker}(d\pi_x)$.
I do not know how to prove the other direction.
I have doubts even about the partial proof above. How do we know about the existence of a smooth curve $\gamma$ on $\tilde{M}_{\pi(x)}$ such that $\gamma(0)=x, \gamma'(0)=v$. How do we construct it?