Proving commutativity of addition for vector spaces

Normally commutativity is taken as an axiom, but you can deduce it from associativity, distributivity and from the existence of inverses as follows:

$(u + v) - (v + u) = (u + v) - v - u$ (by distributivity)
$ = u + (v - v) - u$ (by associativity)
$ = u + 0 - u = (u + 0) - u = u - u = 0$

So $u + v = v + u$

Your first equation essentially answers the question: $$u+v+u+v=2(u+v)=2u+2v=u+u+v+v$$ From here, because we know that a vector space is a group under addition, add on the left by $-u$ and on the right by $-v$ to get $$-u+u+v+u+v-v=-u+u+u+v+v-v$$ $$v+u=u+v$$

This is legitimate because by definition a vector space is a group under addition. If your definition doesn't have this as part of it, I'd recommend adding your definition.