If there is k in I with V(k)=0 then set x_i = a(k,i) and the claim follows.
Otherwise, it is not always true as the following counterexaple shows.
Let V be 3-dimensional with basis e_1,e_2,e_3 and let I be the set with elements 1,2,3.
Set V(j) = K e_j.
Let W be the set of all families a(i,j) as in the problem. Then W is a vector space of dimension 18.
Let U be the set of all families x(i) as in the problem. Then U is a vector space of dimension 9.
The claim would be that the linear map sending x(i) to x(i)-x(j) is surjective, which cannot be by dimension reasons.
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