The f-block valence electrons understandably include the outermost two from 6s. But why do the f-block elements have an oxidation number of +3, pulling an electron from an inner shell, not even a d-block electron?
First of all, lets define which elements we're discussing as f-block. Generally the lanthanides are treated as f-block elements, though this introduces some ambiguity as La (which starts the lanthanides) has 1 d electron and no f electrons while Lu (which ends the series in some definitions) has 1 d-electron and all 14 4f electrons. So one of these two is a d-block element but its not entirely clear which one. Recall that in the f-block there can only be 14 elements due to the 14 electrons that fill f orbitals (just as the d-block can only have 10 elements).
Not all of the lanthanides have 5d electrons, only La, Ce, Gd and Lu do! The rest fill the 6s and 4f only. The 6s are heavily shielded and easily lost. Only one 4f electron is lost because the 4f closer to the nucleus and more effected by losing electrons, ie once one is removed its difficult to remove another because the rest are far more attracted to the nucleus.
So the real question should be why are La, Ce and Gd exceptions to the filling order? For La and Ce this is because they have a lower Z effective, they're not splitting the energy of the 4f and 5d as effectively. For these two the 5d is lower energy and fills first, then the 4f fills. For the later Ln with a higher Z effective the 4f (being in a lower shell) is pulled to significantly lower energies and fills first.
The reasons for Gd having a different filling order is slightly more complex but can be related to whats seen in the transition metals, recall that Rh and Ru don't actually fill the 5s despite not being at a half-shell closure, they fill one 5s and the rest in 4d. Lu is not an exception.