Arrow pushing: why does the O become positive when its lone pair becomes a double bond?
Ah, the confusion that layman’s terms can cause. You are assuming sharing to mean ‘both have the same electrons as before, they just share a little bit of them’ — but that’s not really the nature of a chemical bond let alone the idea behind formal charges. Rather, once a molecule forms the electrons are delocalised across the entire molecule and the Lewis structures we draw are a mere approximation that makes things easier for our brains.

When determining things such as formal charges, you should stick exactly to the book and perform it strictly as directed. To determine a formal charge, all bonds must be formally cleaved homolytically, i.e. oxygen only gets two electrons back from the double bond (and another from the other single bond). Then count the resulting electrons and compare to the initial state. Since you will arrive at five electrons for oxygen (which is one less than six), it gets a single, formal positive charge. And indeed physically, there is less negative charge density on oxygen than there was before. (But note that not all formal charges correspond well with actual charge density.)

This is a distinct process from determining oxidation states, whereby you must cleave the bonds heterolytically with the electronegative atom taking it all.

In none of these formal processes you can say ‘but it was originally oxygen’s electrons …’ You cannot make electrons wear pink and yellow hats (not even during Winterbash) and thus you cannot distinguish them.