So if you've not tried it, do have a go first, before reading on.
Walkthrough
Step 1
The smallest digit is "1", so any arrow line passing through two or more cells must sum to at least 2. Since we've been told that each box contains a different circled digit, the only circle than can contain a "1" is the short arrow extending down from the cell r3c2.
And, now that the only one cell arrow has been taken, the circle containing the digit "2" must be attached to a two cell arrow. The only two digits that can sum to 2 are 1+1. And only one box may contain a repeat of the digit "1", which we have identified as being box 3. These facts, taken together, mean that the circled "2" must be in box 2, since the other three circles are either already claimed or have arrows incapable of being 1+1 due to the pair of 1s already placed in box 3.
Step 2
The circled "2" (which is in box 2 and on row 2) implies box 1 either r2c1 or r2c2 must also contain a "2", and and that box 1 only contains a single "2", which implies that neither r1c1 nor r1c2 can be a "2".
r1c2 may not be a 1, both because there are already a pair of 1s in column 1, and because it would make box 1's circled digit a "2" (which has already been claimed by box 2). r1c2 may not be a 4 because there is also a 1 on that arrow line, which would require the sum to be 5.
Therefore r1c2 is "3"
Step 3
The 1 and 3 on its arrow line make box 1's circled digit the "4", leaving the remaining digit ("3") to go in the remaining box that doesn't have its circled digit set (box 4).
Since we already know r4c2 is a "1", this tells us, by arrow logic, that r3c1 must be a "2", which resolves the remaining two digits in box 1 and identifies it as being the box where "1" is the missing digit.
Step 4
For each digit, there is only one row upon which that digits appear twice (and one of those two appearances must be as the circled digit for that row). Since the circled digit for row 1 is the digit "4", we know that none of the other digits will appear twice on that row. Which means that, since there is a "3" on row 1 in box 1, neither r1c3 nor r1c4 may contain a "3". Therefore r1c3 is a "4", box 2 is the box that does not contain the digit "3", and (by elimination) box 4 must be the box that does not contain the digit "2".
Step 5
Resolve the last couple of cells, then systematically check our solution against the rules.
A different digit in each arrow circle: check
Each box has a different digit missing, and a different digit that appears twice: check
Each row has a different digit missing, and a different digit that appears twice: check
Each column has a different digit missing, and a different digit that appears twice: check
Each box, row and column have exactly 3 cells that show up red if you don't turn off 'show clashes': check
CTC version (solution):
Penpa version (solution)
Note: to trigger the answer check in penpa, you will need to fill in the empty cells of all the tables (sudoku-normal digits in dark-blue ink works, for example; though if you want to use red in the outer tables to indicate missing digits, or green to indicate index number of row/col/box, penpa does permit them to match a blue digit in the answer check)
The full 9x9 puzzle, for which the above is an example, can be found at:
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SudokuSolves/StandardRulesDoNotApplyRulesExample
