(I assume that $A, B, C, D$ represent names, and denote four pairwise distinct players; I call the other four players $E,F,G,H$. Furthermore, I assume that the statement "Since $C$ and $D$ were rivals ..." does not cover the particular game played between $C$ and $D$.)
By the problem statement, no player in $\{A,B,E,F,G,H\}$ has won both games against $C$ and $D$. Hence $\{A,B,E,F,G,H\}$ can be partitioned into three disjoint sets:
- the set $L(C)$ of players who have lost to $C$ and won against $D$
- the set $L(D)$ of players who have lost to $D$ and won against $C$
- the set $L(CD)$ of players who have lost against both $C$ and $D$
By symmetry we now assume that $C$ ended up with at least as many points as $D$.
(1) $C$ scored at least $4$ points.
Proof: Suppose for the sake of contradiction that $C$ scored at most $3$ points.
If $C$ won against $D$, this implies $|L(C)|\le2$ and hence $|L(D)|+|L(CD)|\ge4$; then $D$ has scored at least $4$ points; contradiction.
IF $C$ lost to $D$, this implies $|L(C)|+|L(CD)|\le3$ and hence $|L(D)|\ge3$; then $D$ has scored at least $4$ points; another contradiction.
(2) $A$ and $B$ each scored at least $5$ points.
Proof: $A$ and $B$ won the first place, and hence scored more points than $C$.
(3) $A$ and $B$ each scored at most $5$ points.
Proof: Suppose for the sake of contradiction that $A$ and $B$ both scored at least $6$ points. They played one game against each other, and altogether $12$ games against the other six players. Then one of them must have won all six matches against ${C,D,E,F,G,H}$, and in particular his matches against $C$ and $D$; contradiction.
So the answer is: $A$ and $B$ each lost two matches and each won five matches.