2 made the answer simpler
source | link

May bemy original answer was unnecessarily complicated. A Table command is all what is needed

k = 4;(*must be <= 2*n-1 where n is length of list below*)
list = {x1, x2, x3, x4, x5, x6, x7};
Clear[f]
Flatten@Table[f[Sequence @@ list[[1 ;; (2 i - 1)]]], {i, Range[k]}]

Mathematica graphics

Original

k = 4; (*must be <= 2*n-1 where n is length of list below*)
list = {x1, x2, x3, x4, x5, x6, x7};
Clear[f]
Flatten@MapIndexed[ {z = First[#2];
    f[Sequence @@ list[[1 ;; (2 z - 1)]]]} &, Range[k]]

Mathematica graphics

May be

k = 4; (*must be <= 2*n-1 where n is length of list below*)
list = {x1, x2, x3, x4, x5, x6, x7};
Clear[f]
Flatten@MapIndexed[ {z = First[#2];
    f[Sequence @@ list[[1 ;; (2 z - 1)]]]} &, Range[k]]

Mathematica graphics

my original answer was unnecessarily complicated. A Table command is all what is needed

k = 4;(*must be <= 2*n-1 where n is length of list below*)
list = {x1, x2, x3, x4, x5, x6, x7};
Clear[f]
Flatten@Table[f[Sequence @@ list[[1 ;; (2 i - 1)]]], {i, Range[k]}]

Mathematica graphics

Original

k = 4; (*must be <= 2*n-1 where n is length of list below*)
list = {x1, x2, x3, x4, x5, x6, x7};
Clear[f]
Flatten@MapIndexed[ {z = First[#2];
    f[Sequence @@ list[[1 ;; (2 z - 1)]]]} &, Range[k]]

Mathematica graphics

1
source | link

May be

k = 4; (*must be <= 2*n-1 where n is length of list below*)
list = {x1, x2, x3, x4, x5, x6, x7};
Clear[f]
Flatten@MapIndexed[ {z = First[#2];
    f[Sequence @@ list[[1 ;; (2 z - 1)]]]} &, Range[k]]

Mathematica graphics