I do not think it possible to arrange that all opponents are different, there must be three pairs of players who play together twice - below these are (10 8) (11 9) (12 7):
(12 7 1 8) (14 11 9 5) (6 3 13 10) (4 2)
(10 8 2 9) (14 12 7 6) (4 1 13 11) (5 3)
(11 9 3 7) (14 10 8 4) (5 2 13 12) (6 1)
There is another unavoidable issue, the 6 players who get to play in the two-balls, will of course have 2 fewer opponents than the other 8 players who always play in four-balls.