How to group each element with each consecutive elements in one head

How about a simple table?

Table[f @@ list[[{i, j}]], {i, 4}, {j, i, 4}]

If you want to use this for a general list, you should use Length[list] in the table iterators or maybe:

With[{n = Length[list]},
 Table[f @@ list[[{i, j}]], {i, n}, {j, i, n}]
]

Solutions

---

Pick[
 Outer[f, list, list],
 UpperTriangularize@ConstantArray[True, {#, #} &@Length@list]
 ]

Using the new Composition shorthand:

Thread@*f @@@ MapIndexed[{#, list[[First@#2 ;;]]} &, list]

Timings

---

Testing with

list = Range[1000];

the first method takes 0.363 seconds to complete and the second takes 0.120 to complete. As a comparison, halirutan's Table method took 1.183 to complete. RunnyKine's is the fastest I have tested of the others, taking just 0.336 to complete. All times were measured with AbsoluteTiming.

This is not the shortest, but faster than all except Pickett's (almost just as fast)

f4 = Thread@f[#[[1]], #] & /@ Partition[#, Length@#, 1, {1, 1}, {}] &

OR

 dP = Developer`PartitionMap;

Then:

f5 = dP[Thread@f[#[[1]], #] &, #, Length@#, 1, {1, 1}, {}] &

Timings:

Needs["GeneralUtilities`"]

f1 = With[{n = Length[#]}, Table[f @@ #[[{i, j}]], {i, n}, {j, i, n}]] &; 

f2 = MapIndexed[#[[#2[[1]] ;;]] &, Outer[f, #, #]] &;

f3[x_] := Thread@*f @@@ MapIndexed[{#, x[[First@#2 ;;]]} &, x];

BenchmarkPlot[{f1, f2, f3, f4, f5}, RandomInteger[999, #] &, 2^Range[12