I understand the particular form of vectors of subspace $S$ of Free vector space $F(V \times W)$;
$$ (v+v',w)-(v,w)-(v',w) $$
$$ (v,w+w')-(v,w)-(v,w') $$
$$ a(v,w)-(av,w) $$
$$ a(v,w)-(v,aw) $$
They are just the "right candidates" to force bilinearity. But, it seems that they are really a set of four vectors. I mean, the claim is that they generate the subspace $S$. Hence, we can write a linear combination with then.
So, can I say that every vector of the subspace $S$ is written like: $$u = c^{1}(v+v',w)-(v,w)-(v',w) + c^{2}(v,w+w')-(v,w)-(v,w')+c^{3}a(v,w)-(av,w)+c^{4}a(v,w)-(v,aw) \equiv c^{1}e_1 + c^{2}e_2 + c^{3}e_3 + c^{4}e_4 ?$$
Put in another way, they are a basis of four vectors like: $$\{e_1,e_2,e_3,e_4\} \equiv \{(v+v',w)-(v,w)-(v',w) , (v,w+w')-(v,w)-(v,w') , a(v,w)-(av,w) , a(v,w)-(v,aw)\}?$$