You need to use the definition of a direct sum (i.e. of $\oplus$) here. The definition you use (or a relevant theorem regarding direct sums), you should be able to immediately say
Every $v \in V$ can be uniquely written in the form $v = v_1 + v_2$ with $v_1 \in W_1$ and $v_2 \in W_2$. (In other words: if $v_1 + v_2 = w_1 + w_2$ with $v_j,w_j \in W_j$, then $v_1 = w_1$ and $v_2 = w_2$)
Now, we can define a bilinear form $g$ by
g(v,w) = g(v_1,w_1) + g(v_2,w_2)
where the $v_j,w_j \in W_j$ are chosen so that $v_1 + v_2 = v, w_1 + w_2 = w$, as is guaranteed by the definition of $\oplus$. Because the decomposition is unique, this function is unambiguous. That is, there is only one function that fits this description. It remains to be shown that $g$ is bilinear.
To that end, note that for $v,v' \in V$ and $\alpha \in K$, note that
g(v,w) + \alpha\,g(v',w) =
g_1(v_1,w_1) + \alpha g_1(v_1',w_1) + g_2(v_2,w_2) + \alpha\, g_2(v_2',w_2)
g_1(v_1 + \alpha v_1',w_1) + g_2(v_2 + \alpha v_2',w_2)
Now, here's a tricky point: take $\tilde v = v + \alpha v'$. We see that $\tilde v_1 = v_1 + \alpha v_1'$ and $\tilde v_2 = v_2 + \alpha v_2'$ is such that $\tilde v_1 \in W_1$, $\tilde v_2 \in W_2$, and $\tilde v = \tilde v_1 + \tilde v_2$. It follows from this and the above that
g(v + \alpha v', w) = g(\tilde v,w) =
g_1(\tilde v_1 ,w_2) + g_2(\tilde v_2, w_2)
\\ = g_1(v_1 + \alpha v_1',w_2) + g_2(v_2 + \alpha v_2',w_2)
\\ = g(v,w) + \alpha\,g(v',w)
So, we see that $g$ is indeed linear in the first argument. Similarly, we may argue that $g$ is linear in its second argument, and hence bilinear.