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You can get the row sums using {2} as the second argument of Total:

Total[{{1, 2}, {2, 0}, {3, 4}}, {2}]

{3, 2, 7}

Then you can use Ordering and Sort to get the ordering of sums and the sorted sums, respectively:

Through @ {Ordering, Sort} @ %

{{2, 1, 3}, {2, 3, 7}}

and Transpose the resulting pair of lists:

Transpose @ %

{{2, 2}, {1, 3}, {3, 7}}

Combine the three steps to define a function:

ClearAll[indexedSortedSum]
indexedSortedSum = Transpose[Through@{Ordering, Sort}@Total[#, {2}]] &;

indexedSortedSum @ {{1, 2}, {2, 0}, {3, 4}}

{{2, 2}, {1, 3}, {3, 7}}

Alternatively, you can Apply Plus at Level 1 instead if using Total

ClearAll[indexedSortedSum2]
indexedSortedSum2 = Transpose[Through@{Ordering, Sort}[Plus @@@ #]] &;

indexedSortedSum2 @ {{1, 2}, {2, 0}, {3, 4}}

{{2, 2}, {1, 3}, {3, 7}}