I think a simple cyclic schedule will work for you. Number the 13 players as 0 to 12.
( 0 1 4) v ( 3 5 10)
( 1 2 5) v ( 4 6 11)
( 2 3 6) v ( 5 7 12)
( 3 4 7) v ( 6 8 0)
( 4 5 8) v ( 7 9 1)
( 5 6 9) v ( 8 10 2)
( 6 7 10) v ( 9 11 3)
( 7 8 11) v (10 12 4)
( 8 9 12) v (11 0 5)
( 9 10 0) v (12 1 6)
(10 11 1) v ( 0 2 7)
(11 12 2) v ( 1 3 8)
(12 0 3) v ( 2 4 9)
As you ask, all 12 possible partners occur exactly once over the 6 matches for any one player. The opposing players should occur at least once, and at most twice.