As electronegativity increases, the stability of a negative charge (e.g. pair of electrons) will increase. You can think of going from sp3 to sp2 to sp hybridization as changing the effective electronegativity of the atom.

What’s the opposite of a lone pair of electrons? An empty orbital. As electronegativity increases, the *instability* of an empty orbital increases. That’s why we see carbocations, but almost never see nitrogen, oxygen, and (especially) fluorine with completely empty orbitals.
Applying the analogy above, you can think of going from sp3 to sp2 to sp as increasing the effective electronegativity of the atom, with the result that an empty orbital will be less stable.
Another way to look at it is from the perspective of potential energy. Think of an object 1 km above the surface of the Earth. It has a certain potential energy that is related to the gravitational force of the planet. If you keep that distance constant but increase the mass of the planet (say Earth -> Uranus -> Jupiter) you are increasing the potential energy of that object, just as increasing electronegativity correlates with increased potential energy.
When we’re talking about “destabilization of positive charge” due to hybridization (or electronegativity) you can really think of it as decreasing the ionization energy.

I would not say that H2S is a better base than H2O. I have never seen a pKa value for H3S+

I know aldehydes are more reactive than ketones because they have less CH/electron donating groups to stabilize the partial positive charge on the carbonyl C. But esters have two oxygens bonded to the carbonyl carbon; my mind tells me “Shouldn’t that ester oxygen destabilize/pull more charge away/make the carbonyl C more positive?”. Another part of my mind tells me “An O-CH3 is an electron donating group; shouldn’t it stabilize the partial positive on the carbonyl C?”. But I guess this is another resonance beats induction thing? A ketone only has one resonance structure, placing the positive charge fully on the carbonyl carbon, whereas the ester has two resonance structures, one where a positive charge is placed on the ester oxygen (thus making the positive charge less concentrated).

When in doubt, I always remind myself that “resonance beats everything”, and this is why, for example, halogen substituents on a benzene, while being deactivating through induction, are still ortho/para directors through resonance.

Radicals are neutral because the negative charge from the electrons balances out the positive charge from the nucleus.

That’s a common trap! In the phenyl carbocation [C6H5 (+) ] the empty p orbital is in the same plane with the C-H bonds, which is 90 degrees away from the p orbitals of the aromatic ring. So resonance stabilization is not possible.