Is it because after the monobromination of anisaldehyde the 3-Br-4-MeO-BA is too deactivated or what?
Yes, exactly. After the first bromination, you have one strongly deactivating group (-CHO) and one weakly deactivating group (-Br), while you have only one activating group (-OCH3), this making the whole molecule too deactivated.
Hydroxy is significantly more activating than methoxy, so in the case of 3-Br-4-OH-BA, the molecule is just enough activated to be brominated a second time, while 3-Br-4-MeO-BA is not.