linear algebra - Surjectivity of bilinear forms.

The answer to Question B is no, as I'll show below.

Let $U=V=mathbf{Q}^3$ and $W=mathbf{Q}^4$. Define
$$beta((u_1,u_2,u_3),(v_1,v_2,v_3))=(u_1 v_1,u_2 v_2,u_3 v_3, (u_1+u_2+u_3)(v_1+v_2+v_3)).$$

Claim 1: $beta$ is not surjective.

Proof: In fact, we will show that $(1,1,1,-1)$ is not in the image. If it were, then we would have a solution to
$$(u_1+u_2+u_3)(u_1^{-1}+u_2^{-1}+u_3^{-1})=-1.$$
Clearing denominators leads to an elliptic curve in $mathbf{P}^2$, but MAGMA shows that all its rational points lie in the lines where some coordinate vanishes.

Claim 2: After base extension to any completion $k$ of $mathbf{Q}$, the bilinear map $beta$ becomes surjective.

Proof: Given $(a_1,a_2,a_3,b) in k^4$, we need to show that it is in the image. If $a_1=0$, then set $u_1=0$, $u_2=1$, $u_3=1$, $v_2=a_2$, $v_3=a_3$, and then solve for $v_1$ in the remaining constraint. The same argument works if $a_2=0$ or $a_3=0$. If $a_1,a_2,a_3$ are all nonzero, then we must find a solution to
$$(u_1+u_2+u_3)(a_1 u_1^{-1}+a_2 u_2^{-1}+a_3 u_3^{-1})=b.$$
Clearing denominators leads to the equation of a projective plane curve
with a smooth $k$-point $(1:-1:0)$, so by the implicit function theorem
there exist nearby $k$-points with $u_1,u_2,u_3$ all nonzero,
which gives us the solution we needed. $square$

If you want a reasonable answer to Question A, I'd suggest that you try to make it more focused.