in this reaction, Sn(0) is oxidized to Sn2+, and your organic ArNO2 is reduced to ArNH2. So you end up with a positively charged metallic species, and an organic species bearing an electron-rich nitrogen atom. Lewis acid an Lewis base.
I have the impression that the exact stoichiometry of the tin 'salts' that form is quite difficult to establish; I'd say that you get various complexes where the metal coordinates your amine, chloride anions, water, etc. Although at a very acidic pH your amine might be too strongly protonated to form complexes with tin at all (ArNH3+). But then solubility might come into play. Never mind.
When you add NaOH, the pH increases, and as water is present there will be an increase in the concentration of OH-. These hydroxide anions will gradually displace your amine from the complexes with Sn2+, and once you have something like Sn(OH)2 you just need to have dehydration to go to SnO.
But please note that the exact mechanisms probably are much more complex.
Concerning the boiling step, I suppose it's just a matter of kinetics. A reaction might be thermodynamically favoured but too slow; increasing the temperature usually helps, as you know from Arrhenius' equation.