Why does oxygen in OH have an oxidation state of 2-?
First step: draw the Lewis structure of $\ce{OH-}$ with all lone pairs. You should be getting three lone pairs on oxygen and one $\ce{O-H}$ bond.

Second step: ‘[cleave] [all bonds] heterolytically in such a way that the electronegative partner gets both electrons.’[1] You should now have eight electrons around the oxygen and zero on hydrogen.

Third step: count the electrons on both (or all) atoms and check what the difference is to their elemental valence electron count. Oxygen formally has eight electrons on it which is two more than the six it would in its elemental state. The two extraneous electrons give an oxidation state of $\mathrm{-II}$. Likewise, hydrogen has lost its electron for an oxidation state of $\mathrm{+I}$.

Control step: add up the oxidation states of all atoms. If you did everything correctly, they will add up to the entire charge of the ion. Here, this works out:

$$(-2) + (+1) = -1$$



[1]: https://chemistry.stackexchange.com/a/66242