All such schedules will have a minimum of 6 pairs of competitors who never meet. This is an unavoidable problem that is

discussed here.

So I think the following is about the best you can do:

(10 9 v 4 2) ( 3 12 v 6 8) (11 7 v 5 1)

( 6 10 v 5 3) ( 4 12 v 7 9) (11 8 v 1 2)

( 7 6 v 1 4) ( 5 12 v 8 10) (11 9 v 2 3)

( 8 7 v 2 5) ( 1 12 v 9 6) (11 10 v 3 4)

( 9 8 v 3 1) ( 2 12 v 10 7) (11 6 v 4 5)

Players play in partnership with 5 of the other players exactly once, and in opposition with a different 5 players exactly twice. In other words no pairs of competitors are meeting three times.