$SL_2$ is well-known to have one irreducible representation in each dimension, produced by taking the $k$th symmetric power of the standard representation. The global sections of $\mathcal O(1)$, being just the dual space to $V$, of course form an irreducible representations, and the global sections of $\mathcal O(n)$ are the $n$th symmetric power. Other representations from the meromorphic sections are infinite-dimensional, since a finite-dimensional space of sections can only have poles at finitely many points, but the translates of a section with one pole has poles at infinitely many points.
If you would like to understand the action on the points of a line bundle, it is hopefully sufficient to decompose the line bundle into orbits. $\mathcal O(\pm n)$ always has two orbits: the zero section and everything else. To see this, first note that $SL_2$ acts transitively on $\mathbb P^1$, so you need only understand orbits within a certain fiber. The stabilizer of a point on $\mathbb P^1$ is the group of upper-triangular matrices with determinant $1$. These have an eigenvalue of $\lambda$ on the vector corresponding to that point, so they act like multiplication by $\lambda$ on the projective line bundle and like multiplication by $\lambda^n$ on its $n$th power. This is transitive on the nonzero numbers for $n\neq 0$.
The stabilizer is the group of upper-triangular matrices with determinant $1$ whose eigenvalue is an $n$th root of unity. Thus, the larger orbit is isomorphic to the quotient of $SL_2$ by that subgroup. In the case of the projective line bundle, this is just the quotient of $SL_2$ by the group of upper triangular unipotent matrices.