Yes, the elements of $T_pM$ and $T_qM$ are derivations with different domains, and so $T_pM$ and $T_qM$ are by definition disjoint. This is just a formal property, there is no geometrical reason.
In fact it depends on the construction of tangent spaces whether they are disjoint or not.
An alternative approach is to define $T_pM$ as the vector space of all derivations $D : C^\infty(M) \to \mathbb R$ at $p$. Here $C^\infty(M)$ is the algebra of all $C^\infty$ real-valued functions on $M$, and a derivation at $p$ is a linear map such that $D(f g) = (Df)g(p)+ f (p)Dg$. All such derivations have the same domain, and certainly the zero function belongs to all $T_pM$.
For submanifolds $M$ of $\mathbb R^n$ one can also construct a nice geometric variant of $T_pM$ as a certain linear subspace of $\mathbb R^n$. See for example Why is the tangent bundle defined using a disjoint union? or tangent vector of manifold or Equivalent definition of a tangent space? These "geometric" $T_pM$ are never disjoint. If $M$ is an open subset of an affine subspace of $\mathbb R^n$, then all $T_pM$ are even identical. For $M = S^{n-1} \subset \mathbb R^n$ all antipodal points have identical tangent spaces.
Tu writes
In the definition of the tangent bundle, the union $\bigcup_{p∈M} T_pM$ is (up to notation) the same as the disjoint union $\bigsqcup_{p∈M} T_pM$, since for distinct points $p$ and $q$ in $M$, the tangent spaces $T_pM$ and $T_qM$ are already disjoint.
Formally this is not correct. The disjoint union of an indexed family of sets $X_\alpha$ is defined as
$$\bigsqcup_{\alpha \in A} X_\alpha = \bigcup_{\alpha \in A} X_\alpha\times \{\alpha\} = \{(x, \alpha) \in \left(\bigcup_{\alpha \in A} X_\alpha \right) \times A \mid x \in X_\alpha \} .$$
If we know that the $X_\alpha$ are pairwise disjoint, we could write
$\bigsqcup_{\alpha \in A} X_\alpha$ instead of $\bigcup_{\alpha \in A} X_\alpha$ to indicate this fact, but this would be just a notational convention and not a definition of the concept of disjoint union.
Another way out could be to understand the phrase "disjoint union" as a synonym for "categorical coproduct in the category of sets". If the $X_\alpha$ are pairwise disjoint, then $\bigcup_{\alpha \in A} X_\alpha$ can in fact be used as a coproduct of the $X_\alpha$, but the problem is that coproducts are not unique. The definition is based on a universal property which specifies its purpose allowing to find many individual constructions which result in isomorphic instances of coproducts. Thus we should not say the coproduct, but a coproduct.
