I hope the following works: Let $U, V$ be ultrafilters on $X$ and $Y$ respectively. One side is clear. If there is a bijection $f\colon X\rightarrow Y$ satisfying RK-condition then the restriction of $f$ to any $A\in U$ satisfies RK-condition and it is a bijection.
Now the other side: Assume that there is $A,B$ and a bijection $f\colon A\rightarrow B$ satisying RK-condition as given in the wikipedia page. Hence, we know that $f$ is defined at every point of $U,$ because $f$ satisfies RK-condition by assumption. Hence, we have a function $f\colon X\rightarrow Y$ and we want to show that it is 1-1 and onto. If $f$ is not onto, then $\exists C\subset Y$ such that $f(X)\cap C=\emptyset.$ Then trivially $f^{-1}(C)=\emptyset\in U$ by the RK-condition. This certainly not possible. Hence $f$ is onto.
Now, suppose that $f\colon X\rightarrow Y$ is not 1-1. This means there are $x_1,x_2\in X$ $x_1\neq x_2$ but $f(x_1)=f(x_2).$ Let $K:=\left\{x_1,x_2\right\}\subset X.$ If $K\cap A=\emptyset$ and $K\notin U$ then consider the set $A\cup K\in U$ since $U$ is an ultrafilter. The set $f(A\cup K)=f(A)\cup f(K)$ where $f(K)$ is a one point set. But since $V$ is an ultrafilter and $B\in B$ we necessarily have $f(K)\subset B$ which contradicts to $A\cap K=\emptyset.$ Hence, $A\cap K\neq \emptyset.$ $K$ contains two points. But these two points cannot be in $A$ simultaneously, because $f$ is a bijection on $A$. Therefore, WLOG say $x_1\in A$ and $x_2\notin A.$ But then consider $f(A)\in V.$ $f(A)$ contains $f(x_1).$ By RK-condition again inverse image $f^{-1}f(A)=A$ should contain $x_2$ too. This is again a contradiction.
Hence, $f\colon X\rightarrow Y$ is 1-1 and onto.