I think the simplest motivation is that of vector fields. We want to be able to assign a tangent vector to each point in the manifold, giving us a "field" of vectors on the manifold. That is, $F$ should be some map such that
$$ F(p)\in T_pM $$
for $p\in M$. So, what's wrong with just saying that? Well, nothing really, if you're only interested in the value of vector fields at a point. If you ever want to look beyond singular points, you need some structure connecting your different tangent spaces. For example, to look at continuity of $F$, we need the space of outputs of $F$ to have a topological structure; to look at differentiability, we need a differentiable structure.
So, we need some space which
- contains all the tangent vectors of $M$
- has the same "level" of structure that $M$ has
To satisfy (1), we simply glue together all the tangent spaces by taking a union. Since the tangent spaces are completely disconnected from one another, we can exemplify this fact by using a disjoint union
$$ TM = \coprod\limits_{p\in M} T_pM $$
The rest of the bundle structure is just there to "lift" the structure of $M$ onto $TM$. With that, we can now define vector fields as functions in the usual way
$$ F: M\to TM $$
and we are able to discuss continuity and differentiability to the extent that $M$ admits such properties. However, this definition isn't "complete" because it allows for e.g. attaching a tangent vector from $T_qM$ to $p$, which doesn't fit our idea of a vector field. This gives us another requirement,
- we need a way to determine which point a vector is tangent to
This is the bundle projection map, $\pi: TM\to M$, and so we can add the requirement on $F$ that $\pi(F(p)) = p$ everywhere.
Here, we created a bundle with a base manifold and a vector space at each point, but we could imagine a more general concept of bundle which just has some kind of space $B$ with some other kind of space $F_p B$ at each point $p\in B$. Even in this general setting, we can see the utility of attaching to each point of the base space some element of its attached space. We define a function
$$ \sigma: B\to FB = E $$
with $\pi(\sigma(p)) = p$ as a cross-section (or just section) of the total space $E$.
Applying this terminology to our original example, we can then reconstruct the more terse definition:
$$ \text{a vector field on } M \text{ is a section }\sigma\text{ of the tangent bundle } TM$$
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.
Best Answer
Your derivation is correct.
An informal way of thinking about it: you are solving for $A$ such that $(P + \epsilon X)^2 = (P+\epsilon X)$, where you can treat $\epsilon^2$ as zero. This gives the same equation for the tangent space at $P$.
As noted in the comments, the manifold actually has four components of different dimensions (two of which are just points). Your appeal to the regular value theorem is correct and proves that each component is a manifold (there are two zero-dimensional components and two four-dimensional components).