 Solution

The two pairs of regular functionsdefine maps to the projective line on the loci where their respective coordinates do not vanish simultaneously. Those loci cover $Y$, since simultaneous failure would force $x=y=z=w=0$. On their overlap the equation $xy=zw$ says that the two projective points are equal. They therefore glue to a morphism

Let $[A:B]$ be the homogeneous coordinates on ${\mathbb{P}}_k^1$. The pullback of the hyperplane divisor $V(A)$ has local equation $x$ on the first chart and $z$ on the second. It is therefore exactly $D_Y=V(x,z)$. Compatibility of the pullback of a sheaf of modules with the line bundle associated to a divisor givesThis is the ruling morphism of the punctured three-dimensional affine quadric cone associated with $D_Y$.