The comments of many discussions and general guides seem to have people stuck on this one, so I thought I'd share my own method. I'm sure there's more than one way to solve such puzzles - coming from years of being passionate about Sudoku, I can see how the approach is similar.

Total time spent on the actual solve was less than half an hour.

You are probably serious about analyzing solutions or maybe, so to speak, new to this type of puzzle. I'm not a professional guide writer (nobody is paying me good money for this) and there's no one right answer but I will show you the method I used and won't give unnecessary detail.

Starting grid.

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1a. row I, 1 yellow
wheel 04 cannot have its 2 yellow along the horizontal
1b. row III, 8 blue
both wheels 09 & 11 need to use their 2 blue, else the row will only have 7 blue remaining
1c. column D, 2 blue
wheel 12 cannot have its 3 blue along the vertical

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2. row I, 2 blue
wheel 03 needs to use its 2 blue
therefore wheel 01 cannot use either of its 1 blues

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3. row III, 4 red
wheels 10 & 11 are not enough, with only 2+1=3 or 1+1=2 reds
therefore wheel 09 needs 3 red on the top and wheel 10 cannot use its 2 red

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4. row III, 8 blue
wheels 9 & 11 & 12 = 8 blue
therefore wheel 10 needs its 1 blue on the top

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5. column C, 2 blue
wheel 07 is not enough, with only 1 blue
therefore wheel 11 needs its 2 blue on the left

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6. column C, 4 red
wheels 03 & 15 are not enough, with only 1+1=2 or 1+0=1 reds
therefore wheel 07 needs its 3 reds on the right

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7. column C, 3 yellow 4 red
wheels 03 & 07 = 3 yellow 4 red
therefore wheel 15 cannot have its 1 red on the right or its 1 yellow along the vertical

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8. row IV, 2 yellow
wheel 15 uses its 1 yellow
therefore wheel 14 has to use its 1 yellow and wheel 13 needs its 2 yellow on the bottom

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9. column A, 5 yellow
wheels 09 & 13 = 5 yellow
therefore wheels 01 & 05 needs their 1 and 2 yellows on the right

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10. row I, 1 yellow
wheel 01 = 1 yellow
therefore wheel 02 cannot have its 1 yellow along the horizontal

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11. row I, 2 red
wheel 04 needs its 2 red on the top

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12. column B, 3 blue
wheel 06 is not enough, with only 2 blue
therefore wheel 14 needs its 3 blue on the left and wheel 06 cannot use its 2 blue

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13. column B, 5 red
wheels 10 & 14 = 5 red
therefore wheels 02 & 06 need their 2 & 1 reds on the left

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14. row II, 3 blue
wheels 06 & 07 = 3 blue
therefore wheel 08 needs its 1 blue on the bottom

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15. column D, 1 yellow
wheel 08 = 1 yellow
therefore wheel 12 cannot have its 1 yellow on the right

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16. row IV, 3 red
wheel 14 uses its 1 red
therefore wheel 16 needs its 2 red on the bottom

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Ending grid.