This is more a request for explanations. There's one thing which makes me really mad:
On the one hand, it seems like I do have an intuitive understanding of the Hairy Ball theorem. I imagine an arrow of unit length (that's for simplicity - generally, of any nonzero length, of course) attached to each point of a sphere. I also insist that arrows attached to close points have close directions (usual metric and connection induced from 3d are implied). After few attempts to visualize, I assure myself that such configuration cannot be nowhere pathological.
In other words, it seems that the above described bundle - which could be obtained from the tangent bundle of $S_2$ by restricting the vectors to have unit norm (of course, the vector structure is lost within the fiber) - does not admit a global section.
On the other hand, nothing stops us from considering the $S_2\times U(1)$ trivial $G$-bundle, which certainly has a global section, e.g. $f(p) = \operatorname{const}$.
Moreover, to my understanding, in the second example we don't even need to stress that the section is a group. We can replace $U(1)$ with its torsor $S_1$, and the trivial bundle $S_2\times S_1$ will still admit a global section.
Now, the question. Why exactly does thinking of $S_2$ with $S_1$ circles (in tangent planes) attached to each point fail as an illustration for the $S_2\times S_1$ bundle?
Here's my attempt to answer. Let's go the other way around and start with $S_2 \times S_1$. On each of $S_1$ circles of those are attached to $S_2$ I can choose a point and declare that this is '$0$' (or '$1$', or any other label). Then, I define a global section $f(p)=$'$0$'. I even came up with an illustration. Here it is:
To any point of the ball I attach a circle. I don't draw it's plane as tangent to the sphere - because this is not required! Clearly, if at any point of $S_2$ the arrows look upwards, this defines a global (constant) section. If we now try to make those circles lie in planes tangent to $S_2$, clearly, at certain point we'll encounter a problem - the arrows will no longer be parallel in the usual sense. But this is just a defect of illustration.
Having said so many words, I would like to understand the difference between the two bundles rigorously: \begin{alignedat}{9} E_1 &= \text{Vectors of unit length in } T S_2\\ E_2 &= S_2 \times S_1 \end{alignedat}
I guess, the difference should be in transition functions, since the base spaces and typical fibers are identical.
Can anyone help me to figure out those?
In still other words: what's wrong with representing $E_1$ with the picture above? Of course, it is weird, but we may draw the tangent spaces the way we want. Does it have to do something with connections? I would be surprised if so, since the ability to define a global vector field on the manifold is not related to ability to parallel transport vectors (one doesn't need Christoffels to define the Lie derivative).
The circle bundle of the sphere derived from the tangent bundle isn't $S^2 \times S^1$ topologically; you can think of a normalized tangent vector as a second point on the sphere with a perpendicular vector from the origin, and adding the cross product will yield you a third one and an oriented orthonormal basis. The space of orthonormal bases is homeomorphic to the group $SO(3)$, which is homeomorphic to $\mathbb{RP}^3 \not\cong S^2 \times S^1$.