On the proof that $4$-velocity transforms like vector Let $U$ and $U'$ be the $4$-velocities associated to the coordinates $(t,x)$ and $(t',x')$ related through the Poincaré transformation $P:\mathbb R^4\to\mathbb R^4$, i.e. $(t',x')=P(t,x)$.$^1$
Of course the Jacobian $\Lambda\in\mathbb R^{4\times 4}$ of $P$ is a Lorentz transformation. I extracted the following derivation of the tranformation rule for $U$ from this question:
\begin{equation}
U'=\frac{\mathrm dX'}{\mathrm d\tau}=\frac{\mathrm d(P\circ X)}{\mathrm d\tau}=\Lambda\cdot\frac{\mathrm d X}{\mathrm d\tau}=\Lambda\cdot U\in\mathbb R^{4\times 1}
\end{equation}
As far as I understand, we used the fact that
\begin{equation}
\forall \tau:X'(\tau)=(P\circ X)(\tau),
\end{equation}
but this is not trivial, is it? I will explain my reasoning and I hope for a confirmation/verification:
We have $X(\tau):=X(t(\tau))$, where $I\ni t\mapsto X(t)$ is the $4$-position and $\tau\mapsto t(\tau)$ is the inverse of proper time, i.e. the inverse of the function
$\newcommand{\d}{\mathop{}\!\mathrm{d}}$
\begin{align}
I\ni t\mapsto \tau(t)=\int_{t_0}^t\sqrt{1-\frac{v(\widetilde t)^2}{c^2}}\d\widetilde t+c
\end{align}
for some $t_0\in I$ and $c\in\mathbb R$. So what we are really assuming is the following:
\begin{equation}\tag{1}
X'\circ t'=P\circ X\circ t
\end{equation}
Let $\Pi:\mathbb R^4\to\mathbb R$ be the projection to the time component, then $X'=P\circ X\circ(\Pi\circ P\circ X)^{-1}$ and hence $(1)$ follows from the fact that
\begin{equation}
t'=\Pi\circ P\circ X\circ t
\end{equation}
which is equivalent to
\begin{equation}
\tau=\tau'\circ\Pi\circ P\circ X
\end{equation}
and which can be proven through a change of variables.$^2$ Am I right?

$^1$ The reader familiar with manifolds will note that $(t,x)$ is a chart $\phi: M\to\mathbb R^4$ and that $P=\phi'\circ\phi^{-1}$.
$^2$ We are exploiting the fact that $\tau$ and $\tau'$ are only defined up to a constant when we assume that $t$ and $t'$ have the same domains.
 A: I think you are making this unnecessarily difficult. Here is a proof.
$$
U = \lim_{\delta\tau \rightarrow 0} \frac{X(t+\delta\tau) - X(t)}{\delta\tau}
$$
Now use that $\delta\tau$ is invariant and the difference of two 4-vectors evaluated at a given event is itself a 4-vector (which is easy to prove). It follows that $U$ is a 4-vector multiplied by a scalar invariant, hence it is a 4-vector.
A: As @Frobenius said in the comments, this is a bit too complex to understand a relatively easy equation. I would like to make your equations  bit clearer here.
As you said, $X'(t)=(P\circ X)(\tau)$ should actually be:
$$(X'\circ t')(\tau)=(P\circ X)(\tau)$$
Then, applying inverse function on the right:
$$(X'\circ t' \circ (t')^{-1})(\tau)=(P\circ X \circ (t')^{-1})(\tau)$$
$$(X')(\tau)=(P\circ X \circ (t')^{-1})(\tau)$$
Using the projection operator:
$$(\tau)=(\Pi \circ P\circ X \circ (t')^{-1})(\tau)$$
Why? Because $(X' \circ t')(\tau) \ne (X \circ t')(\tau)$. Therefore applying  $(\Pi \circ X')(\tau) \ne (t')(\tau)$. By the way, in my definition projection operation is something that takes out the first component of a vector. And $X'$ is just a function that is different from $X$.
So I would summarize this redundant set of operations as the following:

*

*Take proper time and apply an inverse transform on it which is just the inverse of the operation that a Lorentz boost would do on the proper time.

*Use this as an argument of your 4-position vector.

*Apply the Lorentz boost.

*Take the projection in time.

Voilà! You are at the very same time where you started.
