Proving both definitions of vertical bundle are the same Let $\pi:E\rightarrow M$ be a vector bundle morphism. The vertical bundle $VE$ is usually introduced as the disjoint union of spaces $V_yE=T_yE_{\pi(y)}$, for $y\in E$. Since $E_p$ can be regarded as a vector space itslef, all the $V_yE$ are naturally identify with $E_p$, proving that have the same diemsnion. But this condition is not enough to ensure $VE$ is a vector subbundle of $TE$. Then one consider the rigurous approach and defines $VE$ as the kernel of $T\pi:TE\rightarrow TM$. However, as far as I kwnow, in this case one has to consider $TE$ as a vector bundle over $M$. Hence, each fibre is now the union
$$
\tilde\pi^{-1}(p)=\bigcup\{T_yE:\pi(y)=(p)\} ,
\tag{$\ast$}
$$
where $\tilde\pi $ is the composition $\pi_{TE}\cdot \pi$. The kernel will now be a subbundle of $TE$, but this time $VE$ will be regarded as a bundle over $M$, while it is usually defined over $E$.
How can I make the link between both definitions? The problem I see is that the theorem we use to ensure the kernel of a vector bundle morphism is a subbundle only considers the case where both are defined over the same manifold.
Thank you
Addendum
Thanks to the conversation kept with @jgon , I have managed to understand why both definitions are equivalent.
I claim that knowing who the fibres are is important and that they change depending on the base manifold we set. It is important to keep this in mind because the kernel depends on the fibres, we consider the kernel of the map $T\pi$ restricted to the fibre of $TE$, if fibres change the kernel also changes. This will not be important in the end, because we can skip from one fibres to others considering or not the union of ($\ast$). 
Now my reasoning. Consider $v\in\tilde\pi^{-1}(p)$. In particular, it will ne in some $T_yE$. Let $c:\mathbb R\rightarrow E$ be a curve representing $v$. Considering a local trivialisation on $E$ and a chart on $M$, we can identify $c(t)$ as $(c_1(t),\dots,c_n(t),c_{n+1}(t),\dots, c_{n+r}(t))$. Then
$$
T\pi(v)= (c_1(t),\dots,c_n(t)),
$$
and $T\pi(v)=0$ implies that the $c_i$ are constant. Then $v\in T_yE_{\pi(y)}$. The converse is also simple, proving that
$$
\operatorname{ker} T\pi|_{\tilde\pi^{-1}(p)} = T\pi(v)=\bigcup\{T_yE_p: \pi(y)=p\}.
$$
To come back and consider $VE$ as vector bundel over $E$, just set $(VE)y=T_yE_p$ instead of the union.
What do you think?
P.S. I have added an addendum instead of deleting the quesion and asking another one, because there were some comments. But if you think it is not the best idea, feel fre to suggest what I should do.
 A: $TE$ is a vector bundle over $M$. It's a vector bundle over $E$, which is itself a vector bundle over $M$, and you can check that an $E$-trivializing coordinate neighborhood of a point in $M$ will trivialize $TE$ over $M$.
Yes, $VE$ will be a bundle over $M$, but it's a subbundle of $TE$, so it's also a bundle over $E$. (Note that we need to use the fact that $\pi$ is a submersion to show that $\ker T\pi$ is a subbundle.)
The first definition uses the fact that $\pi: E\to M$ is a submersion, so all points of $M$ are regular values, thus we can identify the kernel of $T\pi$ at $x\in \pi^{-1}(m)$ with the tangent space of $\pi^{-1}(m)$ at $x$ for any $m\in M$.
Let me try drawing a diagram
$$\require{AMScd}
\begin{CD}
TE @>T\pi>> TM\\
@VVV @VVV\\
E @>\pi>> M
\end{CD} \quad
\begin{CD}
(x,v) @>T\pi>> (\pi x, \pi_{x,*} v)\\
@VVV @VVV\\
x@>\pi>> \pi x
\end{CD}
$$
The kernel of $T\pi$ consists of all pairs $(x,v)$, such that $\pi_{x,*}v=0$. If $U\subseteq M$ is a trivializing neighborhood for $\ker T\pi$, then $\pi^{-1}(U)$ will be a trivializing neighborhood for $\ker \pi$ over $E$.
