I am directly going to address the source of your confusion: $\text{+M}$ of -$\ce{OCH3}$ is more than -$\ce{OH}$. You see, you are basically saying that the $\text{+I}$ effect of $\ce{CH3}$ in -$\ce{OCH3}$ is going to push more electron density towards the ring, hence, -$\ce{OCH3}$ is more activating than -$\ce{OH}$.
But look at the bigger picture, and let's evoke Bent's rule here.
You see, due to steric reasons,the $\ce{C_{(\text{benzene})}-O-C_{(\text{alkyl group})}}$ bond angle of $\ce{OCH3}$ is going to be more than the $\ce{C_{(\text{benzene})}-O-H}$ bond angle of the -$\ce{OH}$. Hence, the percent $\mathrm{s}$-character is going to go up for the central oxygen atom in $\ce{OCH3}$ relatively (to rationalize this, think of the bond angles associated with $\mathrm{sp^3, sp^2,}$ and $\mathrm{sp}$ hybridization and their respective $\mathrm{s}$-characters)
So now, due to increased $\mathrm{s}$-character on both sides of the $\ce{OCH3}$ oxygen, you get an overall more electronegative substituent for the ring as compared to $\ce{OH}$, as the electrons displaced towards the oxygen from the methyl in $\ce{OCH3}$ would still experience a lowering in the energy of the $\sigma_\ce{(O-CH3)}$ bond. Hence, as compared to $\ce{OH}$, the ring will be activated less by $\ce{OCH3}$. I feel that you can now complete the reasoning, by checking the stability of the conjugate bases.
Note: A better qualitative reasoning could have been provided via frontier MO theory, but I chose not to evoke it here. If the OP requests, I can supply the same as well.