1)"Why does withdrawing of electron density from an acidic centre result in the stabilization of the anion formed from dissociation? Or is this a misconception?"
Imagine you're an acid molecule. You can only show your "acid" characteristics if you're able to dissociate your proton (H+) when you're in water. If you lose your proton easily, you're a strong acid.If you can't lose your proton, you are not a good acid(weak acid). This is the background you need to know about acids first.
Next to the question about the importance of stability: After you lose your proton (suppose from your hand), you are now a negatively charged ion floating in water (with a negative charge at your hand). Because there is now a -1 charge in just your hand, there is a dipole between the rest of your body and your hand. This is considered not as "stable" as let's say if the -1 is dispersed throughout your body. The presence of the dipole would "encourage" the proton you lost to return to you. And if it does, you're no longer an acid.
If it requires more work for you to remain safely as a negatively charged ion in water (because of an uneven charge spread), it would be less probable that you exist in that state. Hence, it is said that you are less "stable".I've tried to simplify it.
"Also, what determines whether a group is electron withdrawing or electron donating?"
This is easier to answer. One part, Electronegativity. It is a measure of how strongly an atom pulls electron (cloud) towards itself. Highly electronegative atoms are Fluorine (F), Oxygen (O) and Nitrogen (N) in that order too. F > O > N. you can google "electronegativity chart" and look at images of the values. You can also see how the pKa changes with the atoms near the charge in your picture. (Lower the pKa the stronger the acid, btw).
Carbon and Hydrogen are much less electronegative and they both have similar electronegativity. Hence, compared to F, O, N, there can be considered to be electron donating as they are at the other end of the spectrum.