These $\mathbb{C}$-varieties are both rational threefolds, so they are birationally equivalent.  First, in $Y$ consider the open subset $$D_+(w) = \text{Spec} \ \mathbb{C}[u/w,v/w,r,s]/\langle (u/w)^2 + r(v/w)^2 + s \rangle.$$  The linear projection of $D_+(w)$ to $\mathbb{A}^3$ by $(u/w,v/w,r,s)\mapsto (u/w,v/w,r)$ is an isomorphism.  Thus $Y$ is rational.

Next, in $X$, consider the open subset $U =X \setminus \text{Zero}(x,y)$.  There is an action of $\mathbb{G}_m = \text{Spec}\ \mathbb{C}[t,t^{-1}]$ on $U$ by $t\cdot (x,y,[\lambda,\mu]) = (tx,ty,[\lambda,\mu])$.  This action realizes $U$ as $\mathbb{G}_m$-torsor over the surface $\Sigma = A\times B \cong \mathbb{P}^1 \times \mathbb{P}^1$ with homogeneous coordinates $([\chi,\upsilon],[\lambda,\mu])$ via $(x,y,[\lambda,\mu]))\mapsto ([x,y],[\lambda,\mu])$.  Inside $B$, let $B^o$ be the open subset $D_+(\lambda\mu(\lambda^2+\mu^2))$.  Define $U^o$ to be the open inverse image of $B^o$ under projection.  Consider the morphism $T:U^o\to \mathbb{G}_m$ by $T(x,y,[\lambda,\mu]) = \lambda \mu x/(\lambda^2+\mu^2)$.  It is straightforward to check that $T$ is $G$-invariant, and thus factors as $U^o/G \to \mathbb{G}_m$.  Because of this, $U^o/G$ is isomorphic to $\mathbb{G}_m\times (A\times B^o/G)$.  

Since $A\times B^o/G$ is a unirational surface, it is rational by Lüroth's theorem (it is also not hard to find an explicit rational parameterization).  Therefore, $X$ is also rational.