The finished grid:
The two arrows that share the 2-cell sum have the non-9 digits adding up to at least 21 (1-6) and at most 33 (3-8). So the 2-cell sum is at least half of 21+2×9 = 39 and at most half of 33+2×9=51. Since this sum is an integer, it must be between 20 and 25, inclusive. The second digit cannot be even, since all even digits are marked in the row, and cannot be 21 for the same reason, so it is either 23 or 25.
R2C5 must be 1. 2 is blocked in its column, and if it were 3, then the thermos force the sum of R3C4 and R3C6 to be at least 9, and R4C3 and R4C7 have to sum to at least 3, meaning the sum of both arrows would be at least 18, forcing both circle to be 9. Now looking at the XV clues in box 3, the V must be a 23 pair, since 1 is blocked in the column. Then the V between boxes 1 and 2 must be a 14 pair, since either 2 or 3 is blocked by the other V. The grid thus far:
Let's look at the thermos more carefully:
In row 5 we have 6 digits on the high-end of two thermos, and the right one is already forced to be high. In fact, this lets us fix R5C4: if it is 5 or greater, then row 5 would contain 6 digits that are at least 5, of which there are only 5. Thus R5C4 is 3, forcing the low end of that thermo. We can also conclude that R5C3 can be at most 5 using similar logic. We also remove 2 as a possibility from R9C3, forcing R9C7 to be 2.
Now let's go back to the arrows at bottom. We know the sum is either 23 or 25, so remembering that the 9 gets counted twice in the arrows, the sum of the digits on the arrows is either 2×23-9 = 37 or 2×25-9 = 41. Thus the sum of the digits in the row not on the arrows is either 4 or 8. But for that sum to be 4, we would need the excluded digits to be 1 and 3, but 1 has already been used on the arrows. Hence the sum must be 23. The excluded digits cannot be 17, so they must be either 26 or 35. The grid thus far:
The middle row:
R5C5 must be at least 4, so the only places for 1 and 2 are on the ends, forcing a 12 pair which resolves. R4C3 is at least 3, forcing R5C2 to be at least 6, so we know R5C5 must be either 4 or 5. Now in row 7, there is only one place for the 2, at the left end, which forces 6 to the right. Now, R7C3 and R7C4 must sum to 13, but the only remaining possibility is 58, which resolves the entire row. Some additional Sudoku deductions from there yield the grid:
The arrow at left:
By Sudoku, R4C3 must be at least 3, so R3C4 must be 2, as the alternative is to be at least 6, yielding a sum of at least 10. Thus R4C3 is 456, yielding a 456 triple in column 3. Thus the remaining three cells must be 3/7/9. In particular, at least one of R2C3 or R3C3 is 3/7, meaning the X in box 1 must be either 28 or 46. But in fact it must be 28, since the 2 in column 2 must be in box 1, and has to be on the X. This eliminates 8 from R5C2, which eliminates 5 from R4C3, forcing R5C3 to be 5, and R5C5 4. By Sudoku, we also have R4C6 is 2.
Now focus on the right arrow. If R4C7 is 1, then R3C6 must be 6 or 7, both of which are barred by the column. R4C7 can't be 2,3 or 5 by the column, and can't be 6 or greater by the arrow, so it must be 4. This fixes all of column 7. We can also fix all of row 5 and the arrows up top, and then Sudoku gets us the even digits in the bottom row and all of box 8. The grid thus far:
Column 6 forces R6C6 to be 1, and 5 is also placed in box 5. Pencilmarking box 2 yields a hidden 78 pair in R1C5 and R2C4, forcing R3C5 to be 6, and existing pencilmarks force the 6 in row 2 to be in R2C1. The 59 pair in box 2 also resolves, and we also find a 37 pair in column 3 which forces R4C4 to be 9. The 8 in box 6 looks up to box 2, allowing us to resolve almost all of the top 3 boxes, excepting a 45 pair in R3C1/2.
The 5 placed in box 3 looks down allowing us to finish box 9, and thus box 6 with basic Sudoku. The rest is just cleanup.