= Solution
Every output $u_1+u_2$ lies in the <sum of subspaces> $W_1+W_2$, since it is a <linear combination> of members of the two <vector subspaces>. For $u=(u_1,u_2)$, $v=(v_1,v_2)$ and <scalars> $a,b$,
$$
g(au+bv)=(au_1+bv_1)+(au_2+bv_2)=a(u_1+u_2)+b(v_1+v_2)=ag(u)+bg(v).
$$
Therefore \b[$g$ is a <linear map>]. Every <vector> in $W_1+W_2$ is such an output: group the terms of any <linear combination> from $W_1\cup W_2$ according to their originating <vector subspace>. Thus the <addition map of subspaces> is also <surjective>.
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