Vtyjkyuk

Let $U$ and $V$ be finite dimensional subspaces of W over a field F. Let $B_U$ be a basis for $U$, $B_V$ be a basis for $V$, and $B_U cap V$ be a basis for $U cap V$. Here, $B_U cap V subset B_U$ and $B_U cap V subset B_V$. What I need to do is find a basis for $U + V$ in terms of the bases defined above. From observations I think that a basis for $U + V$ is $B_V cup (B_U - B_U cap V)$. But I am not quite sure how to show this. Also, is it always the case that $B_V cap (B_U - B_U cap V)$ is empty?

Yes your intuition for the basis of $U + V$ is correct!

Since $U cap V$ is a subspace, what you do is first take the basis for $U cap V$. Call this basis at $mathcalB=u_1,u_2,...,u_s$ and so $textdim(U cap V)=s$

Note that $mathcalB$ is a basis for $U cap V$ implies $mathcalB$ is linearly independent in $U$ as well as in $V$

Therefore $mathcalB$ can be extended to a basis of $U$ ,namely, $u_1,u_2,...,u_s,v_1,...,v_t$

Similarly $u_1,u_2,...,u_s,w_1,...,w_u$ is a basis for $V$.

Exercise: Prove $u_1,u_2,...,u_s,v_1,v_2,...,v_t,w_1,w_2,...,w_u$ is a basis for $U+V$

Recall that in general

$$dim (U+V)=dim U + dim V - dim (Ucap V)$$

and that we have

$$B_U + Vsubseteq B_V cup B_U$$

therefore to find a basis for $U + V$, that is a set of linear independent vectors which spans $U+V$, we need to start from the set $B_V cup B_U$ and proceed by elimination.