It took me a few minutes to break down this statement and I have summarized as: To be a strong enough base to deprotonate an acid, the bases' conjugate acid must be stronger than the acid that is being deprotonated.
The base's conjugate acid must be weaker, not stronger, than the acid being deprotonated (which, in this case, is methanol).
Alright, so the easiest way to think of this is conceptually because you can get lost in the numbers really quickly. In order for a base to deprotonate methanol, it must cease to exist as the free base and exist as the B-H complex, in which the base has abstracted a proton from methanol. So under what circumstances will this occur? Well, the B-H complex + methoxide must be more stable than the free base + methanol in order for this to occur.
In other words, the base must be more basic than methanol in order to force methanol to act like an acid and give up its proton. In math terms, this means that the base's pKa must be higher than that of methanol because pKa is a measure of acidity and basicity is the simple inverse of that.
A more nuanced view is given below and you can ignore it if you want. Acidity is a thermodynamic phenomenon. That is, the relative stabilities of the protonated and deprotonated forms determine acidity. So since it's a thermodynamic phenomenon, you can look at bond strength as a surrogate for this idea. All a pKa measures is the "bond strength" of the H-X bond of the acid. So low pKa means more acidic means weaker bond and vice versa. Alright, so what happens in an acid-base reaction? In this case, some base will come in and deprotonate the methanol. This leaves you with a new B-H bond, where B is the base, in place of a broken H-X bond. So how will you know if this will occur? Well, think about thermodynamics. If the new B-H bond formed is stronger than the H-X bond broken, then the reaction is thermodynamically favorable, i.e. the deprotonation of methanol will occur. So then the question is reduced to: how do you tell if the new B-H bond formed is stronger? Well, like I said earlier, pKa here can stand as a surrogate for H-X "bond strength."
So if the pKa of B-H is higher than the pKa of H-X, this means that B-H is less acidic than H-X, or that the B-H bond is stronger or less likely to break. Therefore, you have broken a weaker bond and formed a stronger one, thereby making this an energetically favorable process. Conversely, if the pKa of B-H is lower than the pKa of H-X, that means that B-H is more acidic than H-X and thus B-H would much rather exist in its deprotonated form and
not abstract a proton from the acid.
The more advanced chemists among you might be inclined to look up BDE values for acids and bases and will find that what I suggest here is not necessarily true. That's because the BDEs chemists usually measure are homolytic BDEs and I'm talking about heterolytic cleavage here.
