Tangent bundle: disjoint union

In my book Introduction to Smooth Manifolds, I define $T_pM$ as the set of all linear maps $v\colon C^\infty(M)\to\mathbb R$ that satisfy $$ v(fg) = f(p)vg + g(p)vf $$ for all $f,g\in C^\infty(M)$. Note that the zero map is an element of all these spaces, so $T_pM$ and $T_qM$ are not disjoint for $p\ne q$.

For other definitions of tangent vectors, disjointness might follow automatically. Using the disjoint union in the definition of the bundle is just a handy way to ensure that the vector spaces associated with different points are disjoint regardless of the definition being used.

You are right, the "usual" constructions of $T_p M$ as a set of derivations or a set of equivalence classes of curves produces disjoint tangent spaces at distinct points. Thus you could define $T'M = \bigcup_{p \in M} T_pM$ which is the union of pairwise disjoint vector spaces. I used $T'M$ to distinguish it from $TM = \bigcup_{p \in M} \{p\} \times T_pM$.

Has $T'M$ any benefit in comparison to $TM$? There exists a canonical bijection between both sets, and if we introduce the usual smooth structure on $TM$ and (mutatis mutandis) on $T'M$, then this bijection turns out to be a diffeomorphism.

You might argue that the bundle projection $\pi' : T'M \to M$ has the property that the fiber over $p$ is the "true" tangent space $T_pM$, whereas $\pi : TM \to M$ has the fiber $\{p\} \times T_pM$ which is only a copy of $T_pM$. But as you know, there are various constructions for $T_pM$ which yield formally different vector spaces. Hence it is a rather philosophical question what the true variant should be, and $\{p\} \times T_pM$ has the same right as $T_pM$.

In my opinion it is just a matter of taste if you want to work with $TM$ or $T'M$. The same applies of course to other bundles like tensor bundles, cotangent bundles, etc.

Here is an example where the additional $\{p\}$ is useful. For smooth submanifolds $M \subset \mathbb R^N$ there is a nice geometric construction of the tangent space at $p$: Take all smooth curves $c$ in $\mathbb R^N$ whose image is contained in $M$ and which go through $p$ at some $t_0$. Then the set of all derivatives $c'(t_0)$ forms an $\dim M$-dimensional linear subspace $\tilde T_pM$ of $\mathbb R^N$. These $\tilde T_pM$ are not pairwise disjoint. We need the point $p$ to make them disjoint and then define $$\tilde T M = \bigcup_{p \in M} \{p\} \times \tilde T_pM \subset \mathbb R^{2N}.$$ It turns out that this a smooth submanifold of $\mathbb R^{2N}$ which is diffeomorphic to the abstract $TM$.