The dotted line shows the Fermi level. On average, all states below this line are filled with electrons, and all states above this line are empty. So in the top-left, there are no electrons in the conduction band (above the green line) and no holes in the valence band (below the red line) (all the valence states are filled with electrons).
This is not correct. The Fermi level indicates the level at which a state has a 50% chance of being occupied by an electron or hole. In your diagrams (and very frequently in semiconductors) the Fermi level is within the bandgap where we assume there are no states. But this doesn't mean that there are no electrons in the conduction band or no holes in the valence band.
In your diagrams there are orders of magnitude more holes in the valence band than naturally exist in an undoped semiconductor. You can tell because of how close the Fermi level is to the valence band edge. The number of electrons in the conduction band is substantially reduced following the law of mass action: \$np=n_i^2\$.
I'm not entirely sure how to read these diagrams, but I guess that the gate voltage is positive, since electrons want to move close to the gate (they have lower energy there) and holes want to move away from it. That must mean this shows an N-channel MOSFET.
Yes, the gate voltage is positive, this is an n channel device. You are correct about the gate voltage, although externally applied voltages don't determine the type of device. This would be an n channel device because the semiconductor in the channel region is p doped and presumably the source and drain are n doped.
In the bottom-left image, the gate voltage is stronger, and the conduction band reaches the Fermi level. This means there are now some free electrons on the left side of the channel which can be used to conduct electricity.
Yes, correct. See how even though the Fermi level does not actually cross into the conduction band there is still going to be a substantial inversion charge shown on the right. This is the channel through which current may flow.
If we apply an opposite-polarity voltage to the gate, will the bands bend up?
Yes. The channel region would act like a mos capacitor operating in accumulation. You get extra holes and further remove some of your already small supply of electrons.
And then, can the channel conduct using holes?
Sure, if we ignore the source and drain regions for a minute. Although the semiconductor in the channel region already could conduct using holes without applying any gate voltage since there was already a substantial amount of holes there. After applying a negative gate voltage there are even more.
We know that MOSFETs don't work that way, so I suspect there will be a problem at the drain and source which prevents the MOSFET from conducting.
Yes, exactly. The channel could conduct without inverting it, but there are no carriers injected from the source which will conduct through the channel on account on the p-n junctions between the source and channel and the junction between the drain and channel. Once you invert the channel these junctions are no longer present.