You're on to something when you see that the positive/negative charges are distributed to every other atom.  For an electron-donating substituent, the delocalized electrons add electron density to every other atom.  Therefore, the ortho and para positions gain electron density and are better nucleophiles, while the meta positions do not gain significant electron density.

Explaining why cyclobutadiene is very unstable requires a more complex model than resonance structures.  The effect is one which can be best explained using molecular orbital theory.  Assuming the cyclobutadiene is planar, it should form an aromatic system because the four pi orbitals on each carbon will be conjugated.  By symmetry, these four pi orbitals will form four molecular orbitals (a bonding orbital, two degenerate non-bonding orbitals, and an anti-bonding orbital).  Since each pi orbital contributes one electron, the aromatic system will have four electrons total.  This means two will go into the bonding orbital and the other two will each occupy one of the degenerate nonbonding orbitals.  However, this results in a diradical species (i.e. it has two unpaired electrons).  This configuration is very unstable and will result in cyclobutadiene adopting a distorted geometry so that the pi orbitals are no longer planar and cannot form an anti-aromatic system.

When one carries out a similar analysis with benzene, one finds that there are no non-bonding orbitals and that all electrons are paired and occupy bonding orbitals.  Therefore, benzene is aromatic and stable.

If I recall correctly, all the positions are activated relative to benzene when an inductive donating group is involved, however, the ortho and para positions are much more activated.  It all has to do with how the orbitals align themselves for optimal overlap.

For this image of ressonance to work, the p orbital of oxygen must be in the plane of the rest of the benzen orbitals, right? That means, phenol is planar molecule...