How do I calculate the pH of this buffered solution?
You can think of $\ce{NH4OH}$ as a hydrate: $\ce{NH3_{(aq)}}$; no need for complicated reactions interconverting the two.

Concerning the problem you described, at these concentrations all subtleties are negligible, and you basically only need to calculate the nominal concentrations of $\ce{NH4+}$ and $\ce{NH3}$, and plug them into the Henderson-Hasselbach equation.

You start with $\ce{0.5 M NH3}$ and $\ce{0.5 M NH4Cl}$, and add $\ce{HCl}$ to a total nominal concentration of $\ce{0.15 M}$, without altering the total volume (the question is written ambiguously, but that's what it means, most likely). As $\ce{HCl}$ is a very strong acid, it will convert all $\ce{NH3}$ to $\ce{NH4Cl}$ (actually $\ce{NH4+}$, because we are in solution). Therefore your final situation is: $\ce{(0.5-0.15)=0.35 M NH3}$ and $\ce{(0.5+0.15)=0.65 M NH4+}$. This should give you a pH of about 8.98 (lower than the one you would have without the HCl, consistently with the addition of an acid).

So your main error above was that you dissociated $\ce{HCl}$ (correctly), but then just added the resulting $\ce{H+}$ to the total, forgetting that there was a ton of $\ce{NH3}$ around that would react with it.

Another example is when you add a solution of some particular transition metal salt to a solution of a weak acid (could be iron sulfate + hydrogen sulfide for instance). Despite the fact that the salt solution isn't acidic at all, the pH drops considerably compared to the solution of the weak acid alone, simply because the metal and the anion form a very insoluble salt that precipitates, subtracting weak acid anions from the solution and forcing the previously non-dissociated acid to dissociate.

In general, my advice is stick to what you know, don't go looking for complications: if they did not teach you to do very exact/sophisticated pH calculations, they probably don't expect you to pull them out of a magic hat at the exam. And as my stoichiometry teacher used to say, always write down all species you have in solution.