Finding oxidation state and formal charge is pretty easy. First, draw a lewis structure. Then:
Formal charge (F): F = V - N - B where N = # of valence electrons in lone pairs (nonbonding electrons), V = # of valence electrons and B = # of bonds to other atoms. (sometimes this equation is written as B/2, with B being the number of electrons shared in bonds, and 2 meaning that they are divided between the two atoms; these two ways of expressing F are functionally equilvalent).
Oxidation state (O): O = V - N - 2B - X, where V and N are as before and now B = # of bonds to other atoms
that have lower electronegativity and X = # of bonds to other atoms
that have the same electronegativity.
If you think about these equations, you will see that formal charge treats all bonds as being totally covalent (because all bonding electrons are shared equally between atoms) and oxidation state treats all bonds as being totally ionic (because all bonding electrons are awarded completely to the atom with higher electronegativity), except those between atoms of the same element, which are treated as covalent.
So let's look at formate,
, the simplest carboxylate.
If you draw a Lewis structure, you'll see that the H and C have no non-bonding electrons, the carboxyl oxygen has two bonds (double bond) and two lone pairs and the single-bonded oxygen has three lone pairs and one bond.
This means that the formal charges in formic acid for H, C, =O and -O are 0, 0, 0 and -1, respectively, and the oxidation states are +1, +2, -2, -2. Note in both cases the sum of all formal charges and oxidations states are equivalent. In a neutral molecule this sum is zero; in a charged molecule, like formate, the sum equals the overal molecular charge.
So basically, to find oxidation state: you need to draw a lewis structure, with all bonds and lone pairs. Then consider each bond as independent of all the rest and assign which atom has the high electronegativity. Then the oxidation state is determined by taking the valence electrons and subtracting all the electrons that belong to that atom: each nonbonding electron and two electrons for every bond connecting to an atom of lesser negativity. The carbon in formate, for example, is +2 because the valence electrons equal 4, it has no nonbonding electrons and of the four bonds it has formed, only one (two the hydrogen) is to an atom with lower electronegativity (and zero to other carbons), so only the electrons in the C-H bond belong to carbon. 4 - 0 - 2 - 0 = +2. Compare this to the formal charge on carbon, which is determined assuming that one electron from each bond belong to carbon, because all bonds are considered covalent. Here, carbon has 4 valence electrons, no nonbonding electrons and 1 electron from each of its four bonds = 4 - 0 - 4 = 0.
This approach works generally. Try acetic acid now,
For oxidation states, you'll find that all the 3H's are +1, as before, the methyl carbon is -3, the carbonyl carbon is +3, and the two oxygens are again -2. Overall charge is -1, in agreement with what we expect, which is a nice internal check. Why are the carbons -3 and +3? Well, the methyl carbon is -3 because it has 4 valence electrons, three bonds to hydrogens (in which it is awarded 2 electrons per bond) and one bond to another carbon (in which it is awarded 1 electron, because this bond is treated as covalent). So 4 - 6 - 1 = -3. The carbonyl bond, on the other hand, is +3 because it has 4 valence elctrons, three bonds to oxygens (in which it is awarded zero electrons, because oxygen is MORE electronegative than carbon), and one bond to another carbon (as before, awarded 1 electron), so 4 - 0 - 0 - 1 = +3. From formic acid to acetic acid, the carbonyl carbon has changed oxidation state because in formic acid, it was bonded to a hydrogen (awarded both electrons of the bond), but in acetic acid, it is bonded to a carbon (shares the two electrons equally).
I think from this you'll see that neither oxidation states or formal charges are particularly good representation of actual electron location. While it's true that the two carbons in acetic acid probably have different charge densities, because one is closer to two very electron-withdrawing oxygens (which means the formal charge representation is no good), there's no reason to expect there to be a charge differential of SIX ELECTRONS between them! That's absolutely crazy. I think the formal charge representation is much better in this case, but even it isn't very good. And that's neglecting the fact that neither model includes resonance forms.
Anyway, hope that helps answer your question. Any other questions, just ask.