Let $f$ be a binary relation between sets $A_0$ and $B_0$ and $g$ be a binary relation between sets $A_1$ and $B_1$.
It can be proved that $[f\cap(A_1\times B_1)=g]\wedge [g\cap(A_0\times B_0)=f]$ implies $f=g$.
Really, it follows $g\cap(A_0\times B_0)\cap(A_1\times B_1)=g$. Thus $g\subseteq A_0\times B_0$. Consequantly $g=f$.
Do the following hold?
$[f\cap(A_1\times B_1)\supseteq g]\wedge [g\cap(A_0\times B_0)]\supseteq f$ implies $f=g$.
$[f\cap(A_1\times B_1)\subseteq g]\wedge [g\cap(A_0\times B_0)]\subseteq f$ implies $f=g$.
It is true in general that if $X \subseteq Y \cap Z$, then $X \subseteq Y$ and $X \subseteq Z$. The hypothesis of statement (1) implies that $f \subseteq g$ and $g \subseteq f$, and so $f = g$, as required. So statement (1) is true.
The hypothesis of statement (2) holds for all $f,g$, provided that $A_0 \cap A_1 = \varnothing$ or $B_0 \cap B_1 = \varnothing$, so it doesn't follow in general that $f=g$. So statement (2) is false in general.
In fact, more generally, we have $(A_0 \times B_0) \cap (A_1 \times B_1) = (A_0 \cap A_1) \times (B_0 \cap B_1)$, and so the hypothesis of (2) holds whenever the relations $f$ and $g$ agree on $(A_0 \cap A_1) \times (B_0 \cap B_1)$; but then $f$ and $g$ can say what they like about the elements not in these intersections, so they need not be equal.