Definition of tangent bundle. I have tried to write the construction of the Manifold structure on the Tangent Bundle, $T M$. In writing this I have used Tu's book: An Introduction to Manifold, & Warner's book: Foundations of Differentiable Manifolds & Lie Groups.
Defining the set $T M$:
Let $M$ be a $d$-dim Smooth Manifold. The Tangent Bundle of $M$ is the disjoint union of all the Tangent Spaces of $M:$
$$
T M:= \displaystyle\coprod_{p \in M} T_p M = \displaystyle\bigcup_{p \in M} \{p\} \times T_p M 
$$
Notation:  Each element in $T M$ is denoted by $(p,v)$ such that $v \in T_p M$.
As defined, $T M$ is a set, with no Topology or Manifold structure. Our goal is to make $T M$ a Smooth Manifold.

Topology on $~T M$:
Let $(U,\varphi) = (U,x_1,\ldots,x_d)$ be a co-ordinate chart on $M$.
At a point $p \in U$, a basis for $T_p M$ is
$\left\{\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_d}\right|_p\right\}$,
so a tangent vector $v \in T_p M$ can be expressed uniquely as a linear combination of the basis vectors:
$$
v = \displaystyle\sum_{i=1}^n v(x_i) ~ \left.\frac{\partial}{\partial x_i}\right|_p
\ ~~~~~~~~~~~~~~~~~~~ \ (1)
$$
Let us consider the natural projection map $\pi: T M \to M$ defined as follows:
$$
\pi((p,v)) = p
$$
Define $\tilde{\varphi}:\pi^{-1}(U) \to \varphi(U) \times \mathbb{R}^n$ as follows
$$
\mathbf{(p,v) \mapsto (x_1(p),\ldots,x_d(p),v(x_1),\ldots,v(x_d)) = (\varphi(p),v(x_1),\ldots,v(x_d))} 
$$
Claim: $~\tilde{\varphi}$ is a bijection.

*

*$\tilde{\varphi}$ is injective: Let $(p_1,v_1) ~\& ~(p_2,v_2)$ be two elements in $\pi^{-1}(U)$ with
$ ~\tilde{\varphi}(p_1,v_1) = \tilde{\varphi}(p_2,v_2)$. 
As $\varphi:U \to \mathbb{R}^d$ is an injection, we have $x_i(p_1) \neq x_i(p_2)$ for some $i \in \{1,\ldots,d\} ~\implies ~p_1=p_2$. 
Again if $v_1(x_i) = v_2(x_i)$ for all $i \in \{1,\ldots,d\}$, since $p_1=p_2$, by equation (1) we have $v_1=v_2$. 
Therefore $(p_1,v_1) = (p_2,v_2)$, making $\tilde{\varphi}$ as injection.


*$\tilde{\varphi}$ is surjective: For any element $(\varphi(p),c_1,\ldots,c_d) \in \varphi(U) \times \mathbb{R}^n$ we have $\left(p ,\displaystyle\sum_{i=1}^d c_i ~\left.\frac{\partial}{\partial x_i}\right|_p\right)$ $\in \pi^{-1} (U)$ such that
$$
\tilde{\varphi}
\left(p ,\displaystyle\sum_{i=1}^d c_i ~\left.\frac{\partial}{\partial x_i}\right|_p\right)
= (\varphi(p),c_1,\ldots,c_d)
$$
We can therefore use $\tilde{\varphi}$ to transfer the topology of $\varphi(U) \times \mathbb{R}^n$ to $\pi^{-1}(U)$: A set $A$ in $\pi^{-1}(U)$ is open iff $\tilde{\varphi}(A)$ is open in $\varphi(U) \times \mathbb{R}^n.$
Constructing a topology on $T M:$ Let $\mathfrak{F}$ be the differentiable structure on $M$. We define $\mathfrak{B}$ as follows:
$$
\mathfrak{B} := \displaystyle\bigcup_{U \in \mathfrak{F}} \left\{B ~| ~B ~\text{is open in} ~~\pi^{-1}(U)\right\}
$$
Claim: $\mathfrak{B}$ is a basis for some topology on $T M$

*

*$\mathfrak{B}$ covers $T M:$ Let $(p,v) \in T M$ and $(U,\varphi)$ be a chart around $p \in M$, then, by the definition of $\pi$, we have $(p,v) \in \pi^{-1}(U)$, and as $\pi^{-1}(U) \in \mathfrak{B}$, we have
$$
\bigcup_{B \in \mathfrak{B}} B = T M 
$$


*$A,B \in \mathfrak{B} \implies A \cap B \in \mathfrak{B}:$
Let $A$ be an open subset of $~\pi^{-1}(U)~$ and $B$ be an open subset of $~\pi^{-1}(V)~$, then clearly $A \cap B ~\subset \pi^{-1}(U) \cap \pi^{-1}(V)$. Now we will show that $A \cap B$ is an open subset of $\pi^{-1}(U \cap V)$. Note that
$$
\pi^{-1}(U) \cap \pi^{-1}(V) = \pi^{-1}(U \cap V)
$$
So $\pi^{-1}(U \cap V)$ is a subspace of both $\pi^{-1}(U) ~\& ~\pi^{-1}(V)$. Therefore $A \cap \pi^{-1}(U \cap V)$ & $B \cap \pi^{-1}(U \cap V)~$ are open in $\pi^{-1}(U \cap V)$. Now since $A\cap B ~\subset ~\pi^{-1}(U \cap V)~$ we have
$$
A \cap B = (A \cap \pi^{-1}(U \cap V)) \cap (B \cap \pi^{-1}(U \cap V))
$$
Hence $A \cap B$ is open in $\pi^{-1}(U \cap B)$.
We give the Tangent Bundle the topology generated by $\mathfrak{B}$. This topology is Second Countable & Hausdorff.

Manifold Structure on $T M:$
We show that if $\{(U_{\alpha}, \varphi_{\alpha})\}$ is a Smooth Atlas on $M$, then $\{(\pi^{-1}(U_\alpha), \tilde{\varphi_\alpha})\}$ is a Smooth Atlas on $T M$.
We already have that $T 
 M = \bigcup_{\alpha} \pi^{-1}(U_{\alpha})$.
It remains to check that on $\pi^{-1}(U_{\alpha} \cap U_{\beta})$, $~\tilde{\varphi_{\alpha}}$ and $\tilde{\varphi_{\beta}}$ are $C^{\infty}$ compatible.
Let $(U_\alpha,x_1,\ldots,x_d) ~\& ~(U_\beta,y_1,\ldots,y_d)$ be two co-ordinate charts in $M$.
And denoting $U_{\alpha \beta} := U_{\alpha} \cap U_{\beta}$, we have $\pi^{-1}(U_{\alpha}) \cap \pi^{-1}(U_{\beta}) = \pi^{-1}(U_{\alpha \beta})$.
So for any $p \in U_{\alpha\beta}$ there are two bases for the Tangent Space
$T_p M: \left\{\left.\frac{\partial}{\partial x_i}\right|_p\right\}_{i=1}^d ~\& ~\left\{\left.\frac{\partial}{\partial y_j}\right|_p\right\}_{j=1}^d$. Hence any tangent vector $v \in T_p M$ has two descriptions:
$$
v 
= \sum_{i=1}^d a_i ~\left.\frac{\partial}{\partial x_i}\right|_p 
= \sum_{i=1}^d b_i ~\left.\frac{\partial}{\partial y_j}\right|_p
\ ~~~~~~~~~~~~~~~~ \ (2)
$$
Claim: $\tilde{\varphi_\beta} \circ \tilde{\varphi_\alpha}^{-1}: \varphi^{-1}_\alpha(U_{\alpha \beta}) \times \mathbb{R}^n \to \varphi_\beta(U_{\alpha \beta}) \times \mathbb{R}^n$ is a $C^{\infty} map.$
The map is defined in the following way
$$
(x,a_1,\ldots,a_d) 
\mapsto 
\left(
\varphi^{-1}_\alpha(x),
\sum_{i=1}^d a_i ~\left.\frac{\partial}{\partial x_i}\right|_p
\right) 
\mapsto 
\left(
\varphi_\beta \circ \varphi^{-1}_\alpha(x), 
\sum_{i=1}^d a_i ~\left.\frac{\partial y_1}{\partial x_i}\right|_p,
\ldots, 
\sum_{i=1}^d a_i ~\left.\frac{\partial y_d}{\partial x_i}\right|_p \right)
$$
$$
=
\left(
\varphi_\beta \circ \varphi^{-1}_\alpha(x), b_1, \ldots, b_d 
\right)
\ ~~~~~~~~~ \ \text{(Using (2))}
$$
Now as $\varphi_\beta \circ \varphi_\alpha^{-1}$ and the change of coordinates are smooth maps. We have $\tilde{\varphi_\beta} \circ \tilde{\varphi_\alpha}^{-1}$ is a smooth map.
Similarly $\tilde{\varphi_\alpha} \circ \tilde{\varphi_\beta}^{-1}$ is also a Smooth Map.

I have a question, which I am stating below:

*

*While showing that $\tilde{\varphi_\beta}\circ \tilde{\varphi_\alpha}^{-1}$ is smooth, one needs to show that $~\left.\frac{\partial y_j}{\partial x_i}\right|_p$ are $C^{\infty}$ maps. My question is why are $~\left.\frac{\partial y_j}{\partial x_i}\right|_p, ~C^{\infty}$ maps, for all $j \in \{1,\ldots,d\}$?

 A: I will write what you have in a slightly different way that might help.
Let $U \subset M$ be open. Suppose $O_x \subset \mathbb{R}^n$, $n = \dim(M)$ is open and $\phi : O \to U$ is a coordinate chart. Then the natural chart for $TU \subset TM$ is none other than $d\phi : TO \simeq O \times \mathbb{R}^n \to TU$ defined by $d\phi(x)v = (\phi(x), d\phi(x)v)$. Suppose $\psi : \Omega_y \to U$ is another chart. Let $F : \Omega \to O$ be the transition map from $\psi$ to $\phi$, i.e. $F = \phi^{-1} \circ \psi$, or "$x = F(y)$". Then the transition map from $d\psi$ to $d\phi$ is $DF : T\Omega \to TO$, given by
$$DF(y)w = (F(y), DF(y)w).$$
By definition of smooth manifold, $F$ is $C^{\infty}$. So $DF$ is then clearly $C^{\infty}$. Since $F = (x_1, \dots, x_n)$, $\frac{\partial x_j}{\partial y_k}$ is $C^{\infty}$ as well.
A: This is actually just a matter of a very basic definition: what is a smooth map between two open sets in $\mathbb{R}^d$.
Typically, if $U$ is an open set in $\mathbb{R}^d$, a function $f:U\to \mathbb{R}$ is defined as nothing but the partial derivatives
$$
\frac{\partial^{\alpha_1+\cdots+\alpha_d} f}{\partial^{\alpha_1}x_1\partial^{\alpha_2}{x_2}\cdots\partial^{\alpha_d}{x_d}} \tag{*}
$$
are well-defined, finite and continuous for all $(x_1,\cdots, x_d)$ in $U$ and all $a_1,\cdots, a_d \in \mathbb{N}$.
By this definition, you can easily prove that if $f$ is smooth, then all its partial derivatives in the form $(*)$. Especially,
$$
\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2},\cdots \frac{\partial f}{\partial x_d}
$$
are all smooth.
And for two opensets $U, V\subseteq \mathbb{R}^d$, a smooth map $F:U\to V$ is defined as that the $d$ component functions
$$
\begin{aligned}
F_1(x_1,x_2,\cdots, x_d)\\ 
F_2(x_1,x_2,\cdots, x_d)\\
\cdots\quad \cdots \quad \quad\\
F_d(x_1,x_2,\cdots, x_d)
\end{aligned}
$$
are all smooth functions.
For the problem you asked, the map $\tilde{\varphi}_\beta \circ{\tilde{\varphi_\alpha}}^{-1}$ is a smooth mapping from open set $\tilde\varphi_{\alpha}(U\cap V)$ to open set $\tilde\varphi_{\beta}(U\cap V)$, which means exactly that its component expressions
$$
\begin{aligned}
y_1(x_1,x_2,\cdots, x_d)\\ 
y_2(x_1,x_2,\cdots, x_d)\\
\cdots\quad \cdots \quad \quad\\
y_d(x_1,x_2,\cdots, x_d)
\end{aligned}
$$
are smooth functions, therefore, according to what I have mentioned before, we have that the $n^2$ functions
$$
\frac{\partial y_{i}}{\partial x_1}, \frac{\partial y_{i}}{\partial x_2}, \cdots, \frac{\partial y_{i}}{\partial x_d},\quad i = 1,2,\cdots, d
$$
are all smooth, or in other words, each of the components of the Jacobian matrix of $\tilde{\varphi}_\beta \circ{\tilde{\varphi_\alpha}}^{-1}$ is a smooth function on the open set $\tilde\varphi_{\alpha}(U\cap V)$ of $\mathbb{R}^n$. This is what you want to proof.
