Proof that two charts on the tangent bundle $TM$ are $C^\infty(M)$-compatible I'm struggling to understand a proof in the "Construction of the tangent bundle" section of the lecture notes downloadable here https://mathswithphysics.blogspot.com/2016/07/lectures-on-geometric-anatomy-of.html (Frederic Schuller's Lectures on the Geometric Anatomy of Theoretical Physics), on page 86 of the document, regarding the fact that one can construct a smooth atlas on the tangent bundle.
In particular, two charts $(preim_{\pi}(U),\xi)$ and $(preim_{\pi}(\tilde{U}),\tilde{\xi})$ have been constructed on the tangent bundle $TM$ from two charts on the base manifold $M$ denoted $(U,x)$ and $(\tilde{U},\tilde{x})$. I understand that we need to show that the chart transition map $\tilde{\xi}\circ\xi^{-1}$ is smooth, but I don't understand why (as it is written in the notes) $$\tilde{\xi}\circ\xi^{-1} : x(U\cap \tilde{U})\times \mathbb{R}^{dimM} \longrightarrow \tilde{x}(U\cap \tilde{U})\times \mathbb{R}^{dimM}$$ has the domain and target that it does. In my head, the map should just go from $\xi(preim_{\pi}(U))\rightarrow\tilde{\xi}(preim_{\pi}(\tilde{U}))$. Why does it instead go from $x(U\cap \tilde{U})\times \mathbb{R}^{dimM} \rightarrow \tilde{x}(U\cap \tilde{U})\times \mathbb{R}^{dimM}$?
 A: To avoid typing too much I'll just write $\pi^{-1}[U]$ instead of $\text{preim}_{\pi}(U)$. First, recall how the chart $(\pi^{-1}[U], \xi)$ is constructed from $(U,x)$: we define $\xi: \pi^{-1}[U]\to \xi[\pi^{-1}(U)]\subset \Bbb{R}^n \times \Bbb{R}^n$ (where $n := \dim M$) according to
\begin{align}
\xi(X) &:= \left((x^1\circ \pi)(X), \dots (x^n\circ \pi)(X), (dx^1)_{\pi(X)}(X), \dots  , (dx^n)_{\pi(X)}(X)\right)
\end{align}
In words, what we're doing is we first take the vector $X \in \pi^{-1}[U]$. Now note that this vector lies in a particular tangent space, namely $X\in T_{\pi(X)}M$; so the base point is $\pi(X)$. So there are two pieces of information we have to keep track of, the fist is the base point, and the second is the actual "vectorness" aspect of it, which is why the first $n$ entries are $(x^i\circ\pi)(X)$, which keep track of the chart representative of the base point, while the second $n$ entries $(dx^i)_{\pi(X)}(X)$ keep track of the components of $X$ relative to the chart induced basis.
Now what you can show is that the image of $\xi$, namely $\xi[\pi^{-1}(U)]$ equals exactly $x[U]\times \Bbb{R}^n$. How do we show this? Well, it's very simple, note that by construction, the following set inclusion is "obvious"
\begin{align}
\xi[\pi^{-1}(U)]\subset x[U]\times \Bbb{R}^n
\end{align}
For the reverse inclusion, note that if $(a,v) = (a^1, \dots, a^n, v^1, \dots, v^n)\in x[U]\times \Bbb{R}^n$, then
\begin{align}
X:= \sum_{i=1}^n a^i \dfrac{\partial}{\partial x^i}\bigg|_{x^{-1}(a)}
\end{align}
is a vector which lies in $T_{x^{-1}(a)}M$; and since $\pi(X) = x^{-1}(a) \in U$, this means exactly that $X\in \pi^{-1}(U)$. Also, it is easy to see from the definition of $\xi$ that $\xi(X) = (a,v)$.
What we have just shown is that for every $(a,v) \in x[U]\times \Bbb{R}^n$, there exists an $X \in \pi^{-1}(U)$ such that $\xi(X) = (a,v)$. This is exactly what it means to prove $x[U]\times \Bbb{R}^n \subset \xi[\pi^{-1}(U)]$. Thus, these two sets are actually equal.

Finally, we apply this to your actual question. The domain of the transition map $\tilde{\xi}\circ \xi^{-1}$ should actually be $\xi[\pi^{-1}(U) \cap \pi^{-1}(\tilde{U})]$. Now, we're just going to apply a few simple set theoretic identities:
\begin{align}
\xi[\pi^{-1}(U) \cap \pi^{-1}(\tilde{U})] &= \xi[\pi^{-1}(U \cap \tilde{U})] = x[U \cap \tilde{U}] \times \Bbb{R}^n.
\end{align}
A similar reasoning shows that the target space is $\tilde{\xi}[\pi^{-1}(U)\cap \pi^{-1}(\tilde{U})] = \tilde{x}[U\cap \tilde{U}]\times \Bbb{R}^n$.
