theChemcoach, the hydronium ion is H3O+. It doesn't really matter whether you use the H+ or H3O+ notation, though in general H3O+ is better and more accurate. Also, dissolution is always the conversion of a substance from it's state at STP (or whatever it's initial conditions, generally the former), to the aqueous state. Here, dissolution is represented by the eqn HCl (g) -> HCl(aq), and this simply means that the dipole-dipole forces amongst HCl and the hydrogen bonding network in water are of similar strength and dipole-dipole forces can be formed between solute and solvent. This is distinct from the deprotonation step, which is (in terms of Lowry-Bronsted acids/bases): HCl(aq) + H2O(l) --> H3O+(aq) + Cl- (aq). This reaction only happens to occur because it is favourable (Ka large) due to thermodynamic and kinetic considerations.
As for electronegativity, it can basically be conceived of as the ability of an atom to attract electrons in a bond towards itself. In other words, it arises mainly from the electrostatic interaction between the positively-charged, shielded nucleus of an atom and the outermost electrons of the adjacent atom in the bond. i.e. the greater the number of protons in the nucleus for comparable atomic radii, the EN will be larger, or for comparable numbers of protons in the nuclei, the EN will depend on the atomic radius and number of electrons shielding the nucleus from other atoms. In our cases (O, F, Cl), the oxygen and fluorine atoms have similar radii, and the fluorine atom has 1 more proton, which outweighs the effect of the added, shielding valence electron, and so F's EN will tend to be greater than that of oxygen (4.0 vs 3.5). In comparing the oxygen atom with chlorine, the fact that the oxygen atom is so tightly bound vs the chlorine atom's 2 electron shells outweighs the fact that chlorine has more nucleons, and protons, in its nucleus, as it will have a greater number of shielding electrons in addition.