Why in the union $\bigcup_{p \in U}T_p^*(\mathbb{R}^n)$ all of the sets $T_p^*(\mathbb{R}^n)$ are disjoint? I am currently reading the An Introduction to Manifolds by Loring W.Tu (2nd edition, pp. 34), and as a novice to differential geometry and topology it is not quite obvious to me why in the union $\bigcup_{p \in U}T_p^*(\mathbb{R}^n)$ all of the sets $T_p^*(\mathbb{R}^n)$ are disjoint?
Namely, let $U$ be an open subset of $\mathbb{R}^n$ and denote the cotangent space to $\mathbb{R}^n$ at $p \in \mathbb{R}^n$ by $T_p^*(\mathbb{R}^n)$. Then $T_p^*(\mathbb{R}^n)$ is (by my understanding) the set of all linear mappings from the tangent space $T_p(\mathbb{R}^n)$ to $\mathbb{R}$ at $p$. So far the author has not defined precisely what a tangent space is, but I have understood it to be a vector space of all tangent vectors at a given point $p$, where the space is spanned by partial derivatives of all the basis vectors of the surrounding space: $\frac{\partial}{\partial x_i}\vert_p$.
With this in mind, does the disjointedness stem from a set theoretical definition of a mapping $f$ being the set of all ordered pairs $(x, f(x))$? If so, then why cannot we have two tangent spaces that are equal for different points $p, q$, if all that matters is the partial derivatives by the spanning vectors of the surrounding space?
 A: Unfortunately Tu has not given a precise definition of the tangent space $T_p(\mathbb R^n)$ when he introduced $T^*_p(\mathbb R^n)$ - and this is a source of confusion.
The tangent space $T_pM$ is properly introduced in Chapter 3 "The Tangent Space". Tu defines it as the set of all derivations $d : C^\infty_p(M) \to \mathbb R$. Here $C^\infty_p(M)$ is the algebra of germs of $C^\infty$ real-valued functions at $p \in M$. These algebras are pairwise disjoint for the points $p \in M$, thus also the $T_pM$ are pairwise disjoint and so are their dual spaces.
But let us come back to Tu's preliminary definition of $T_p(\mathbb R^n)$ on p. 10.

In calculus we visualize the tangent space $T_p(\mathbb R^n)$ at $p$ in $\mathbb R^n$ as the vector space of all arrows emanating from $p$. By the correspondence between arrows and column vectors, the vector space $\mathbb R^n$ can be identified with this column space. To distinguish
between points and vectors, we write a point in $\mathbb R^n$ as $p = (p_1, . . . , p_n)$ and a vector in the tangent space $T_p(\mathbb R^n)$ as
$$v = \left[ \begin{array}{rrr}
v_1  \\ 
. \phantom{.} \\
. \phantom{.} \\
. \phantom{.} \\
v_n  \\ 
\end{array}\right] \phantom{xxx} \text{or}  \phantom{xxx}  \langle v_1,\ldots,v_n\rangle .$$
....
Elements of $T_p(\mathbb R^n)$ are called tangent vectors (or simply vectors) at $p$ in $\mathbb R^n$ . We sometimes drop the parentheses and write $T_p\mathbb R^n$  for $T_p(\mathbb R^n)$.

To be honest, this is extremely unclear. Does he mean $T_p(\mathbb R^n) = \mathbb R^n$, differing perhaps in notation by using tuples and column vectors? I do not think so. Tu speaks about the vector space of all arrows emanating from $p$, thus it should be interpreted as $T_p(\mathbb R^n) = \{ (p,v) \mid v \in \mathbb R^n \} = \{p\} \times \mathbb R^n$. These are again pairwise disjoint and so are their dual spaces.
Note that Tu explains on p. 11 that tangent vectors $v$ give us directional derivatives $D_v$ which prepares the abstract definition in Chapter 3.
You may like to have a look also at
How Can the Vector Space $\mathbb{R}^n$ be identified with the Column space
Directional derivatives at $P$ are all derivations at $P$
Proof of Isomorphism between Tangent Space and the Vector Space of all Derivations
Why is the tangent bundle defined using a disjoint union?
A: Partial derivatives are by nature a (very) local object, i.e. saying something about the limit of the difference quotient towards a point $p$. On a level of manifolds $M$, the tangent space $T_pM$ tells you, colloquially, in which direction you need to go to stay on the manifold and this may differ from point to point. Also since $T_pM$ is an 'abstract' vector space, which is not naturally embedded in $\mathbb{R}^n$, there is no direct way of comparing different tangent spaces. Both are reasons to define the tangent bundle $TM$ as disjoint union of tangent spaces.
$\mathbb{R}^n$ in that regard is a less furtunate example towards all arguments. We have a globally comparable way of applying partial derivates, at every point in $\mathbb{R}^n$ one may go into 'every direction' and the tangent space is due to $\mathbb{R}^n$ being a vector space naturally embedded in $\mathbb{R}^n$ which makes tangent spaces at different points comparable.
