You don’t have any reason to think that $x$ and $y$ span $V$: all you know is that they are linearly independent. The dimension of $V$ might well be greater than $2$, in which case no two-element subset of $V$ will span $V$, though many will be linearly independent.
You have two things to show:
- If $ad-bc\ne0$, then $u$ and $v$ are linearly independent.
- If $u$ and $v$ are linearly independent, then $ad-bc\ne0$.
It’s probably easiest to prove (1) by proving the contrapositive: if $u$ and $v$ are linearly dependent, then $ad-bc=0$. That’s because the assumption of linear dependence gives you something very concrete to work with: if $u$ and $v$ are linearly dependent, there are scalars $\alpha$ and $\beta$, at least one of which is non-zero, such that $\alpha u+\beta v=0$. Now write this out in terms of $x$ and $y$:
$\alpha(ax+by)+\beta(cx+dy)=0\;.$
Collect the $x$ and $y$ terms on the lefthand side:
$(\alpha a+\beta c)x+(\alpha b+\beta d)y=0\;.$
By hypothesis $x$ and $y$ are linearly independent, so
$\left\{\begin{align*} &\alpha a+\beta c=0\\ &\alpha b+\beta d=0\;. \end{align*}\right.$
This says that $\begin{bmatrix}\alpha\\\beta\end{bmatrix}$ is a non-zero solution to the equation
$\begin{bmatrix}a&b\\c&d\end{bmatrix}z=0\;.$
what does that tell you about $\det\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and hence about $ad-bc$?
To prove (2), again go for the contrapositive: if $ad-bc=0$, then $u$ and $v$ are linearly dependent. You can pretty much just reverse the reasoning in the argument that I outlined for (1).