The curves $A$ and $B$ on $Y=\mathbb P^1\times \mathbb P^1$are bihomogeneous of bidegree $(2,1)$ and $(1,2)$ respectively.
This just means that $A$, say, has degree $2$ in $w$ and degree $1$ in $u$.
The intersection number of curves of bidegree $(m,n)$ and $(m',n')$ is $mn'+m'n$, so it is $2.2+1.1=5$ for $A$ and $B$.
An amusing consequence is that $A$ and $B$ are not linearly equivalent because intersection numbers are compatible with linear equivalence and if we had $A\cong B$ we would get $(A,B)=2.1+1.2=4$ for the intersection number.
You can read about what precedes in Chapter I of Beauville's book Complex algebraic Surfaces.
In the case at hand everything can be calculated explicitly:
The intersection points are $\infty=([0:1],[0:1]) $ and the four points of intersection in the affine open subset $U=\mathbb A^1\times \mathbb A^1$ given by $u_0=w_0=1,u_1=u,w_1=w$ of the curves $A_0=A\cap U$ and $B_0=B\cap U$.
The equations of these curves are $u+w^2=0, \:u^2+w=0$ and the curves are transverse at all four intersection points : this is obvious for the point $u=0,v=0$ and very easy for the other three points after a choice of coordinates putting these points at $(0,0)$.
Similarly $A$ and $B$ are also transverse at $\infty$.