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}]

As expected sum has 25 terms.

We get the same result using Total:

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,

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}]

As expected sum has 25 terms.

We get the same result using Total:

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,

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}]

As expected sum has 25 terms.

We get the same result using Total:

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

sum == total

True

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

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}]

As expected sum has 25 terms.

We get the same result using Total:

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,