So if $$[a_\textbf{q}^\dagger, a_\textbf{p}^\dagger] = 0$$ and therefore $$a_\textbf{p}^\dagger a_\textbf{q}^\dagger|0\rangle=a_\textbf{q}^\dagger a_\textbf{p}^\dagger|0\rangle$$

means that under particle exchange the two-particle state stays unchanged or $$|\bf p,q\rangle=|\bf q,p\rangle$$ if the two particle state is represented by $|\bf p,q\rangle$. This is not the case if the particles are fermions, since there is a sign change under particle exchange  $$|\bf p,q\rangle=-|\bf q,p\rangle$$ and the creation operators anti-commute.

Note that the statement "_Moreover, a single mode p can contain arbitrarily many particles_" is consistent for bosons and not fermions, since there is no limit to particle numbers with bosons but there obviously is with fermions due to the Pauli exclusion principle.

>I don't see how the statements of this paragraph fit with the fact that for a two-particle Bosonic (or, Fermionic) system, if we swap the particles the wave function does not (or, does) change.

This statement is true for bosons and not fermions for which the state changes sign under the exchange of particles.

He has basically stated that particles obeying the Klein-Gordon equation (the KG equation describes spin zero bosons), or "Klein-Gordon particles" will obey Bose-Einstein statistics using a mathematical argument explaining that the state describing the system does not change under particle exchange.