The bundle corresponding to $0\in \mathbb{Z}$ is, not surprisingly, the trivial bundle.
For $z\in \mathbb{Z}$, the bundle corresponding to $-z$ is the same bundle with opposite orientation on the fibers, so we may as well just try to understand $z = 1, 2, 3,...$. Let $E_z$ be the rank 2 oriented vector bundle corresponding to the positive integer $z$.
Now, stick an inner product on $E_z$ and let $M_z\subseteq E_z$ be the set of all unit length vectors. Note that $M_z = f^{-1}(1)$ where $f:E_z\rightarrow\mathbb{R}$ is $f(p) = |p| $ and that $1$ is clearly a regular value. Hence, $M_z$ is actually an embedded submanifold. $M_z$ is called the unit sphere bundle over $S^2$.
The projection map, when restricted to $M_z$, gives a bundle $S^1\rightarrow M_z\rightarrow S^2$.
Via some abstract nonsense, the number $z$ can be identified with the Euler class of this bundle. The Euler class is technically an element of $H^2(S^2;\mathbb{Z})$, but since we're bluring the distinction between positive and negative numbers, we can take it to be a well defined integer.
There is something called the Gysin sequence which is an exact sequence involving $S^2$ and $M_z$ which can be computed with knowledge of the Euler class. Suffice it to say that we can show, using the Gysin sequence, that $H^2(M_z) = \mathbb{Z}/z\mathbb{Z}$. This is how we tell the bundles apart. In particular, if we can construct a bundle where $H^2(M_z) = \mathbb{Z}/z\mathbb{Z}$, then we must have found them all.
Now, to your questions.
The Euler class of the tangent bundle of an oriented manifold can be identified with the Euler characteristic times the fundamental class. Since the Euler characteristic of $S^2$ is $2$, the tangent bundle is $E_2$ (or $E_{-2}$). The cotangent bundle of a manifold is always bundle isomorhpic to the tangent bundle via the flat/sharp maps (assuming a background Riemannian metric).
As to what the bundle "looks like", here's how I think about it.
Start with the Hopf bundle $S^1\rightarrow S^3\rightarrow S^2$. This is a principal bundle, so there is a "preferred" direction on the $S^1$s. Now, think of each $S^1$ as sitting inside of an $\mathbb{R}^2$, with each $\mathbb{R}^2$ disjoint from every other $\mathbb{R}^2$. Give each $\mathbb{R}^2$ the orientation which makes the "preferred" direction on the $S^1$, say, counter clockwise. This is $E_1$. (Notice that in this case, $M_1 = S^3$ and we have $H^2(S^3) = 0$ as claimed.)
Now, there is a canonical $\mathbb{Z}/z\mathbb{Z}$ inside of $S^1$, the $z$th roots of unity. Dividing by this action, we get a bundle $S^1/(\mathbb{Z}/z\mathbb{Z})\rightarrow S^3/(\mathbb{Z}/z\mathbb{Z})\rightarrow S^2$.
The fiber is still a circle, but the total space becomes a lens space (or $\mathbb{R}P^3$ when $z=2$). Do the same trick of filling in each circle fiber with an oriented $\mathbb{R}^2$ to make the bundle $E_z$. In this case, the sphere bundle is the lens space.
Incidentally, this provides one of the more complicated proofs I know of that the unit tangent bundle of $S^2$ (i.e., $M_2$) is diffeomorphic to $\mathbb{R}P^3$.