What is gained by insisting on the distinction between tangent spaces at different points and double-tagging them in the construction of the tangent bundle? What specifically am I missing by neglecting this distinction?
Definition (tangent space and tangent bundle): Let $M$ be a topological $n$-manifold. For each point $p\in M$, the tangent space at $p$ is $\Bbb R^n$. The tangent bundle on $M$ is the space $TM=M\times\Bbb R^n$.
Motivation/Explanation:
There are four things to keep in mind here
Thus spoke the gods: "the tangent space exists at every point and is defined completely independently of coordinates"
homeomorphism $\not\implies$ diffeomorphism
coordinate independent $\implies$ chart independent $\implies$ atlas independent $\implies$ smooth structure independent
Occam's razor
Why "topological manifold" is sufficient
In order for the definition of the tangent space at a point to be completely independent of the choice of coordinates, it must also be independent of the choice of smooth structure. While we could insist that the tangent space is only defined given a smooth structure, the only sensible way to do this without inducing coordinate-dependence is to tag the tangent space at each point with the entire set of admissible smooth structures - i.e. $T_pM=S\times\mathcal T_pM$, where $S$ is the set of all possible smooth structures on $M$ (considered as a topological manifold), and $\mathcal T_pM$ is the "standard" tangent space. This way, the absence of a smooth structure forces the nonexistence of the tangent space at every point.
However, if we do this, then we concede the necessary distinction between isomorphism and equality - since our ability to assert that $T_pM$ is distinct from $\mathcal T_pM$, is entirely contingent on their being non-equal, despite being isomorphic. For the sake of consistency, then, we ought to employ this convention in general.
That is, given two distinct constructions of the reals $\Bbb R_1\ne\Bbb R_2\cong\Bbb R_1$, the $S^2$ whose charts map to $\Bbb R_1$ is not the same as the $S^2$ whose charts map to $\Bbb R_2$. If we refuse to make this distinction, then equality becomes subjective - which is bad. Is this pedantic? Yes. Yes it is. But is also necessary for internal consistency.
Given that the distinctness of non-equal constructions of the same space (note the conventional use of the word "same") is not usually acknowledged, it seems reasonable that the tangent space at a point should not depend on the existence of a smooth structure. While we could still force dependence in far more elaborate ways, this would increase complexity without providing any real benefit.
So, the most straightforward way to define the tangent space is $T_pM:=V(p,M)\cong\Bbb R^{\dim M}$, where $V$ is some vector space whose definition depends at most on the point $p$ and the topology on $M$ (since, again, a change in smooth structure must yield the same tangent space to retain coordinate-independence). Being that this is a topological invariant, it naturally extends to arbitrary topological manifolds.
Why distinct points have identical tangent spaces
To force the distinctness of identical tangent spaces ("identical" according to the same convention by which $\Bbb R_1=\Bbb R_2$), we must again tag each of the tangent spaces with a point. That is, for $p\in M$, we have something like $T_pM=\{p\}\times V_p$. Why shouldn't we do this?
Well consider that the tangent space is not generally regarded as an affine space. If I were to say "the $T_pM=p+V$, where $V$ is the set of tangent vectors," most users of this site would disagree. But the affine space $p+V$ is distinguished from $V$ itself only by the point $p$. With this in mind, it becomes very difficult to justify the distinction between tangent spaces at distinct points without effectively resorting to "because I said so." And while, again, we can force this distinction by introducing caveats or additional structure, this once more only adds complexity without providing any real benefits.
Once more, the simplest solution is the best: $T_pM=\Bbb R^{\dim M}$.
Why the tangent bundle is a product
In keeping with the above, if we define the tangent bundle so that it can be constructed from tangent spaces, we cannot expect the tangent space to transfer any smooth structure to the tangent bundle - or rather, the smooth structure conferred to the tangent bundle will not necessarily result in $TM$ being a smooth manifold. In this case, however we define the tangent bundle, it will only be defined up to homeomorphism, not diffeomorphism (to do otherwise brings us into conflict with coordinate independence once more).
Now, I could be wrong here - I probably am - but it is my understanding that the topology on $TM$ is that of $M\times T_pM=M\times\Bbb R^{\dim M}$. Now, if there is additional smooth structure atop this, it will depend on the smooth structure on $M$, which, tracing our dependencies, makes the tangent-bundle coordinate dependent. This is not necessarily a deal-breaker - we still have the required coordinate independence of the tangent space - but it does mean that we cannot construct the tangent bundle directly from the tangent space at each point.
So, whether or not $T(S^2)=S^2\times\Bbb R^2$ at this point depends on whether or not the tangent bundle is a topological invariant in the same way that the tangent space is.
