imines

Reductive amination is a two step process. First, an imine or enamine is formed from a carbonyl and an imine.

Then, a the imine (or enamine) is reduced to an amine (using reducing agents such as H₂, NaBH₄, NaBH₃(CN), NaBH(OAc)₃, etc.). For practice on forming imine and enamines, see problem 712).

What this problem is really asking is "Which nitrogen on semicarbazide is more nucleophilic?"

The nitrogens bonded to the carbonyl on semicarbazide all have resonance forms, so their lone pairs aren't as available for nucleophilic attack.

The other nitrogen (on the left) is the most nucleophilic because it isn't involved in any resonance forms, so its lone pair isn't shared with another atoms.

For both imines and enamines, the sp² carbon comes from the carbonyl, and the nitrogen comes from the amine.

This reaction takes place under acidic conditions, so oxygen should never be negative (only neutral or positively charged). 

Acidic mechanisms tend to be a bit of pain, because you have to include many proton transfer steps (protonation and deprotonation).