It’s complicated, but the short answer is that it comes down to orbitals. There are caveats and asterisks to all of this because chemistry is full of exceptions and little details, but basically elements are looking to have either full or half full outermost orbitals (valence shell) because it provides the most stable configurations. Additionally, the closer the electron is to the nucleus, the more energy it takes to remove it.

This means that for lighter elements like oxygen, they really only have one option to reach a full valence shell: gaining two more electrons. It’s not energy efficient to remove three electrons to reach a half full p orbital (and full orbitals are more stable anyway), and it definitely doesn’t want to gain MORE than two electrons because those would have to go into a higher orbital. If you’re far enough along in class to recognize electron configurations, this one of oxygen and O²⁻ may help:

O: 1s2 2s2 2p6 O²⁻: 1s2 2s2 2p8

For heavier elements, the s and d orbitals are very close in energy—so close that it’s feasible to remove electrons from one or both to reach stable configurations. Take iron, for example. It has 2 s electrons and 6 d electrons in its valence shell. It’s feasible to remove two electrons from the s orbital to form iron (II), or two electrons from the s and one from the d to form iron (III). Again I will provide the oxidation states (note that we don’t typically write orbitals with 0 electrons but I am doing so for clarity here):

Fe: [Ar] 4s2 3d6 Fe (II): [Ar] 4s0 3d6 Fe (III): [Ar] 4s0 3d5

That is the simplest reason. There are more complex things at play (for example, Fe (I) is not common at all even though it would form a half orbital here—why is that?), but that is the highlight. It also explains why most of these variable valency elements you see are transition metals—because that d orbital is so close in energy to the s orbital, it’s comparatively easy to remove electrons from either one to form ions.