Suppose $M \to S^2$ is an orientable circle bundle with Euler number $n$. Regarding $S^2$ as $D^2_1 \cup_{\text{id}} D^2_2$, since $M$ is trivial over $D^2_i$ ($i=1,2$), we have $M=(S^1\times D^2_1)\cup_{\partial} (S^1\times D^2_2)$. So we can think $M\to S^2$ as the "union" of two maps $S^1\times D^2_1\to D^2_1$ and $S^1\times D^2_2\to D^2_2$. It is clear that we can assume that $S^1\times D^2_1\to D^2_1$ is just the projection onto the second factor. Assuming this I want to get an explicit formula of the second map $S^1\times D^2_2\to D^2_2$.
I proceeded as follows. The total space of orientable circle bundle over $S^2$ with Euler number $n$ can be obtained from $S^3$ by a Dehn surgery on a unknot with coefficient $n$. So $M=(S^1\times D^2_1)\cup_{\partial} (S^1\times D^2_2)$ and where we may assume that the boundary homeomorphism $S^1\times \partial D^2_2\to S^1\times \partial D^2_1$ is given by $(z,w)\mapsto (w,z^{-1}w^n)$. Then it seeems natural that the map $S^1\times D^2_2\to D^2_2$ should be given by $(z,w)\mapsto z^{-1}w^n$, but then this is not a bundle because the fiber over $0$ is a multiple fiber. What went wrong?