Note that 286 is Binomial[13,3]. This is not a coincidence. The latter is the number of ways one can place 3 sticks around or between 10 stones (a classic combinatorial problem). One gets from each such placement a partition by taking the first element with multiplicity equal to number of stones before (to the left of) the first stick (this number could be zero), the second element is repeated the number of stones between stick 1 and stick 2, with the third element repeated the number of stones between stick 2 and stick 3 and the fourth element repeated the number of stones after stick 3, Apologies for that belabored explanation

So here is some code. First get the stick placements.

Length[tt = Flatten[Table[{i, j, k}, {k, 3, 13}, {j, 2, k - 1}, {i, 1, j - 1}], 2]]
Short[tt]

(* Out[30]= 286

{{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 5}, << 277 >>, {8,
   12, 13}, {9, 12, 13}, {10, 12, 13}, {11, 12, 13}} *)

Now code to transform a placement into a partition of {0,1,2,3} into an ordered set of 10 elements.

partition[oll_List] := Module[{ll = Flatten[{0, oll, 14}], diffs},
  diffs = Differences[ll] - 1;
  Join @@ Table[ConstantArray[j - 1, diffs[[j]]], {j, Length[diffs]}]]

tuples = Map[partition, tt];
Short[tuples, 12]

(* {{3,3,3,3,3,3,3,3,3,3},{2,3,3,3,3,3,3,3,3,3},{1,3,3,3,3,3,3,3,3,3},{0,3,3,3,3,3,3,3,3,3},
  {2,2,3,3,3,3,3,3,3,3},{1,2,3,3,3,3,3,3,3,3},{0,2,3,3,3,3,3,3,3,3},{1,1,3,3,3,3,3,3,3,3},
  {0,1,3,3,3,3,3,3,3,3},{0,0,3,3,3,3,3,3,3,3},{2,2,2,3,3,3,3,3,3,3},{1,2,2,3,3,3,3,3,3,3},<<263>>,
  {1,1,1,1,1,1,1,1,1,1},{0,1,1,1,1,1,1,1,1,1},{0,0,1,1,1,1,1,1,1,1},{0,0,0,1,1,1,1,1,1,1},
  {0,0,0,0,1,1,1,1,1,1},{0,0,0,0,0,1,1,1,1,1},{0,0,0,0,0,0,1,1,1,1},{0,0,0,0,0,0,0,1,1,1},
  {0,0,0,0,0,0,0,0,1,1},{0,0,0,0,0,0,0,0,0,1},{0,0,0,0,0,0,0,0,0,0}} *)

Add spice, adjust to taste.

Here is more general code.

makeTable[size_Integer, pcount_Integer] := Module[
  {ivars, c, iters},
  ivars = Array[c, pcount - 1];
  iters = 
   Thread[{ivars, Range[Length[ivars], 1, -1], 
     Join[{size + pcount - 1}, Most[ivars] - 1]}];
  Flatten[Table[Reverse@ivars, Evaluate@Apply[Sequence, iters]], 
   pcount - 2]
  ]

partition[oll_List, vals_List] := Module[
   {ll = Flatten[{0, oll, 14}], diffs},
   diffs = Differences[ll] - 1;
   Join @@ 
    Table[ConstantArray[vals[[j]], diffs[[j]]], {j, Length[diffs]}]] /;
   Length[vals] == Length[oll] + 1

Test:

t5 = makeTable[8, 5];
tuples = Map[partition[#, {a, b, c, d, e}] &, t5];
Length[t5]
Binomial[8 + 4, 4]

(* Out[155]= 495

Out[156]= 495 *)

In[158]:= tuples[[1 ;; -1 ;; 40]]

(* Out[158]= {{e, e, e, e, e, e, e, e, e}, {b, c, d, d, e, e, e, e, 
  e}, {c, c, c, d, d, e, e, e, e}, {b, b, b, b, b, e, e, e, e}, {a, a,
   a, a, d, d, e, e, e}, {a, a, a, b, b, c, e, e, e}, {b, b, b, b, d, 
  d, d, e, e}, {a, a, a, a, c, c, d, e, e}, {a, a, a, a, a, b, c, e, 
  e}, {b, b, b, b, d, d, d, d, e}, {a, a, a, a, c, c, d, d, e}, {a, a,
   a, a, a, b, c, d, e}, {a, a, b, b, b, b, b, c, e}} *)

Thanks Henrik, that is certainly better than my code and slightly faster.

Paul:

I'm confused. After removing the dupes of d, you have a list of 286 lists that each have 10 integers between 0 and 3.

The integer partitions of with arguments of [10, 4] is a list of 23 lists of lengths between 1 and 4 with values between 1 and 10.

Why are you trying to map one to the other?

If you could provide some context for this question, that might help the community help you.