Here's how I think about it. (Ryan Budney posted his answer while I was typing this. One can think of this as a fleshing out of his answer).

First, we need to understand the unit tangent bundle $T^1 S^2$. Once we know this, we product this with $[0,\infty)$ and then quotient all points of the form $(u, 0)\in T^1S^2 \times\mathbb{R}$ somehow to get the 0 section $S^2$. (This is precisely the mapping cylinder construction Ryan mentions).

Before we can talk about the "unit tangent bundle", we must have a notion of length of vectors. So in the background, pretend like I picked a Riemannian metric so lengths make sense.

I claim $T^1 S^2$ is diffeomorphic to $SO(3)$ (the collection of 3 x 3 orthogonal matrices of determinant 1) which is diffeomorphic to $\mathbb{R}P^3$.

The map from $T^1 S^2$ to $SO(3)$ sends $(u,v)$ to the matrix with columns $u, v, u\times v$. Here, I'm thinking of a unit tangent vector $v\in T_u S^2$ as a vector in $\mathbb{R}^3$ orthogonal to the vector $u$.

The easiest way to see $SO(3)$ and $\mathbb{R}P^3$ are diffeomorphic is to note they are both quotients of $S^3 = SU(2)$ by the same quotienting map.

So, we understand $T^1 S^2$, the unit length vectors in $TS^2$.

To allow for length, we product with $[0,\infty)$. Now, the only problem is the 0 section should be an $S^2$, and it's currently an $\mathbb{R}P^3$, so some quotienting must happen.

What quotienting must happen? Well, all the unit vectors at a given point must collapse to the point. Well, there is the action of a circle on $T^1S^2$ given by rotation vectors clockwise (say) as seen from the normal vector to the sphere. This action is clearly free. Now, it's a fact that if you translate this circle action into the $SO(3)$ picture, the circle action is the Hopf action. This implies that we identify $\mathbb{R}P^3$ with $S^2$ by quotienting by the Hopf action: Two points in $\mathbb{R}P^3$ iff they are in the same Hopf orbit.

Incidentally, I just learned a few days ago that $TS^2$ is not homeomorphic to $S^2\times \mathbb{R}^2$, though I'm still not sure how to prove it ;-). (Of course, it's clear that they are not bundle isomorphic, but they could still be abstractly homeomorphic). I don't know about the other nonparallizable tangent bundles, though.