UPDATE: It's a full solution now.
The first two steps: that the two thermometer tips must be 6,7,8,9 and that the base below the 6 must be 1-5 look fine to me but we can improve the next steps:
Let us put the thermometer base which contains 1-5 on the left:
123 .tt T??
..4 ... t..
..5 6t. ttt
First thing to note is in the top row the 6 can only go to the right-most box.
This in turn forces the following placement of 5s:
123 5tt t..
..4 ... 5..
..5 6t. ttt
Now there is only one square left for the center box 4 and that also determines the right thermometer base:
123 5tt t..
..4 ... 5..
..5 6t4 321
Taking advantage of rotational symmetry we now know that both thermometer bases must be 1,2,3.
But we can do better than that:
Assuming that the 1-5 base is actually the top leads to a contradiction because
..1 ... AAB
..2 ... ABB
543 ... XXX
XXX must sum to 12 which is only possible via 1,2,9 which leaves only 6-8 for the middle box which cannot be arranged in a legal way.
Hence the 1-5 base is the bottom:
345 ... GGF GGF = 19 = 2+8+9
2.. 3.. HHI HHH = 1+3+5
1.. ... HII III = 4+6+7
GGF must sum to 19 which leaves only 2,8,9. This forces HHH = 1,3,5 and the 3 must be the bottom H.
Next noting III = 4+6+7 and focusing on the right-most column we find that FFF = 22 can only be written 5+8+9 or 6+7+9 but as III cannot keep both 6 and 7 off the right-most column only 5+8+9 is possible
..1 ... AAB
..2 ... ABB
t53 ... CCD
t.4 ... CDD
t.t ... EEF FFF = 22 = 5+8+9
5.6 ... EGF GGG = 13 = 2+8+3
345 ... GGF GGF = 19 = 2+8+9
2.. 3.. HHI HH = 1+5
1.. ... 3II III = 4+6+7
Staying in the right-most column we next find that one of the two B's must be 3.
As BBB has to sum to 15 the other two B's must be 4+8 or 5+7.
In either case 1 and 2 must go to the two D's leaving 6 for the third.
..1 ... AAB
..2 ... ABB BBB = 3 + 12
t53 ... CCD
t.4 ... CDD DDD = 6 + 1+2
t.t ... EE5
5.6 ... E38
345 ... GG9 GG = 2+8
2.. 3.. HHI HH = 1+5
1.. ... 3II III = 4+6+7
Next, there is only one way left to make EEE = 15 and that is 9+4+2.
As one of the thermometer tips in row 5 must be 9, so must be the bottom E.
..1 ... AAB
..2 ... ABB BBB = 3 + 12
t53 ... CCD
t.4 ... C6D DD = 1+2
t.t ... EE5 EEE = 15 = 9+2+4
5.6 ... E38
345 ... GG9 GG = 2+8
2.. 3.. HHI HH = 1+5
1.. ... 3II III = 4+6+7
Now the bottom tip of CCC can only be 7 and that almost finalizes the thermometers.
To finalize CCC let us have a look at row 7. Its center block must accomodate 1+6+7, so 1 has to go in the middle.
The two C's in row 3 have to sum to 10, therefore CC = 6+4 or CC = 9+1.
But 6+4 would force the last thermometer square to 7 leaving 1,8,9 for the middle block and, again, 1 would have to go in the middle which is not available.
Therefore CC = 1+9.
..1 ... AAB
..2 ... ABB BBB = 3 + 12
t53 ... CC2 CC = 1+9
8.4 ... 761
9.7 ... EE5 EE = 2+4
5.6 ... 938
345 .1. GG9 GG = 2+8
2.. 3.. HHI HH = 1+5
1.. ... 3II III = 4+6+7
One of the A's in column 7 has to be 6. As 5 and 7 are already present in column 7, the remaining AA = 12 must split 4+8 and the BB = 12 must split 5+7.
..1 ... AAB AAA = 6 + 12
..2 ... ABB BBB = 3 + 12
t53 ... 192
8.4 ... 761
9.7 ... EE5 EE = 2+4
5.6 ... 938
345 .1. GG9 GG = 2+8
2.. 3.. 51I
1.. ... 3II III = 4+6+7
At this point we can do a number of easy steps leading to
..1 ... AAB AAA = 4+6+8
..2 ... A5B BB = 3+7
t53 ... 192
8.4 ... 761
9.7 ... EE5 EE = 2+4
5.6 ... 938
345 .1. GG9 GG = 2+8
27. 3.4 516
16. ... 374
Because of the 6 and 7 in row 7 the 7 in row 6 must be in the center. This forces two more 4's. Now the positions of the 6's in rows 5 and 7 force the 6 in the top center box to be in the middle column. Therefore the 6 in row 3 must be on the thermometer.
..1 ... AAB AAA = 4+6+8
..2 ... A5B BB = 3+7
653 .4. 192
8.4 ... 761
9.7 ... EE5 EE = 2+4
5.6 47. 938
345 .1. GG9 GG = 2+8
27. 3.4 516
16. ... 374
The positions of the 8's in rows 3 and 5 force the 9 in row 4 into the middle.
As positions left and right of the center square have to be 6 and 8 the center square itself must be 3.
This allows a few easy followup moves.
..1 ... AAB AAA = 4+6+8
..2 1.. A5B BB = 3+7
653 .4. 192
834 .9. 761
917 .3. EE5 EE = 2+4
526 471 938
345 .1. GG9 GG = 2+8
279 384 516
168 ... 374
Next, the 2 in row 1 is either in column 5 or 6. Either way the 3 in the top center box cannot be in row 1. After placing the 3 at row 2, column 6 everything falls into place.
781 529 643
492 163 857
653 847 192
834 295 761
917 638 425
526 471 938
345 716 289
279 384 516
168 952 374