Charge operator action in QFT

A quick answer. You should think of $\psi^\dagger$ as the creation operator for a particle, and conversely you can think of $\psi$ as the creation operator of an anti-particle.

Thus since the commutation relation has different sign for $[Q,\psi]$ and $[Q,\psi^\dagger]$, $Q$ will count the particles as positive charge, and anti-particles as negative charge.

The counting would work like this: Assume you have state $|\psi\rangle=\psi^\dagger|0\rangle$ where $|0\rangle$ is the vacuum (i.e. no particles).

The action of $Q$ on this state is

$$ Q|\psi\rangle=Q\psi^\dagger|0\rangle=(\psi^\dagger+\psi^\dagger Q)|0\rangle=\psi^\dagger|0\rangle=|\psi\rangle $$
where we assumes that $Q$ annihilates the vacuums state. Similarly for an antiparticle state $|\bar\psi\rangle=\psi|0\rangle$ we have

$$ Q|\bar\psi\rangle=Q\psi|0\rangle=(-\psi+\psi Q)|0\rangle=-\psi|0\rangle=-|\bar\psi\rangle $$

So we see that $Q$ has positive eigenvalues for particles and negative eigenvalues for anti-particles. Thus $Q$ measures the total charge $Q=N_\psi - N_{\bar\psi}$.