complicatedFunctionOfIndices = FOO[FromDigits @ Sort @ #, FromDigits @ ReverseSort @#2]&

    SeedRandom[1]
    myg = RandomChoice[{3000, 1, 1, 1, 1, 1} -> Range[0, 5], {4, 4, 4, 4, 4, 4}];

    Dimensions[myg]
>     {4, 4, 4, 4, 4, 4}

    sa = SparseArray[myg];

    nonZeroPositions = sa["NonzeroPositions"]
>     {{1, 1, 3, 3, 4, 3}, {1, 4, 3, 3, 4, 4}, {2, 2, 1, 2, 2, 4}, 
>      {4, 2, 1, 2, 1, 4}, {4, 2, 2, 1, 3, 2}}

    nonZeroValues = sa["NonzeroValues"]
>      {3, 5, 3, 1, 4}

    sum = Sum[sa[[## & @@ i]] sa[[## & @@ j]] complicatedFunctionOfIndices[i, j], 
      {i, nonZeroPositions}, {j, nonZeroPositions}]
[![enter image description here][1]][1]

As expected `sum` has 25 terms. 

We get the same result using [`Total`](https://reference.wolfram.com/language/ref/Total.html):

    total = Total[sa[[## & @@ #]] sa[[## & @@ #2]] complicatedFunctionOfIndices[##]  & @@@ 
      Tuples[nonZeroPositions, 2]];

    sum == total
>     True


**Note:** Per Greg Hurt's comment above, replace `"NonzeroPositions"` with `"ExplicitPositions"` if you have version 13.0, 

  [1]: https://i.sstatic.net/qRR3Y.png