Reaction mechanism of 2-amino-4,6-dimethylpyrimidine as a nucleophile
Your mechanism is almost correct; this is essentially an aldol condensation. The only issue is with the protonation states, which (as you've drawn) aren't entirely feasible.

Let's start by establishing how acidic that methyl group actually is. You're right that it is more acidic than a typical methyl group, because the charge can be delocalised onto those two nitrogens. But note that this delocalisation is pretty similar to that in a typical ketone enolate:



so to a very, very rough first approximation, that proton you're removing has a $\mathrm{p}K_\mathrm{a}$ of $\sim 20$, roughly that of a ketone. Yes, there are two nitrogens in the ring which improves the delocalisation, but there is also an amino group sticking off the ring, which is also trying to delocalise electron density onto the ring nitrogens. The value of $4.8$ cited in the comments is far more likely to be the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid (where the amino group is protonated).

Secondly, if anything were to be deprotonated, the $\ce{NH}$ in the amino group is going to be more acidic than the $\ce{CH}$ in the methyl groups. This is exactly analogous to how ammonia is more acidic than methane, and how amide $\ce{NH}$'s are more acidic than ketone $\ce{CH}$'s.

And lastly, note that your reaction is being carried out using acetic acid as a solvent. The chances that a carbanion (and a strongly basic one, as previously established) can exist for any reasonable amount of time under these conditions are... low, shall we say!

Now, you've probably drawn mechanisms for aldol condensations in acid. In this case, you don't deprotonate to form an enolate; you just use the enol tautomer, which is still nucleophilic on carbon. So the strategy here is not to deprotonate the methyl group outright, but to use the tautomer. Basically, use the same mechanism you've drawn, but add a proton to everything:*



That includes the final elimination, too. Note that if you protonate the $\ce{OH}$ group and let it fall off, you can get a benzylic cation that is stabilised by the 4-hydroxy substituent.



* You could delocalise all the way onto the other ring nitrogen if you wanted.