I found a set of rules to solve the problem (TWO + TWO = FOUR) which apply to all the possible results you can find. This does not mean that every time you use these rules a correct result will be found, 22 of the 100 times you can use these rules to find the result, the answer is found, and fortunately there are only 22 results to this problem. At the end of this solution description I will write out all the possibilities.
The first rule is that every possibility comes from adding a number to itself that has a "units" column which is twice the value of its "hundredths" column, e.g. 428 - 8 (units column) is twice the value of 4 (hundredths column). When you reach numbers that when added to themselves reach a number of 1000 or more you simply take the last number of the 2 digit result and replace your "O" with it, e.g. 6 x 2 (of, for example, 632) = 12. From this you take the 2 of the 12 and place it in the units column.
The second rule is that the first rule applies to all numbers where the "tenths" column is a number from 1 - 4, e.g. 614 or 135. When you have a number which is 5 or more in the tenths column, you need to add one to the "units" column, e.g. from 764 to 765.
If these two rules are used when searching for the results you should get the following results: 132 + 132 = 0264, 142 + 142 = 0284, 173 + 173 = 0346, 193 + 193 = 0386, 214 + 214 = 0428, 234 + 234 = 0468, 336 + 336 = 0672, 346 + 346 = 0692, 357 + 357 = 0714, 418 + 418 = 0836, 428 + 428 = 0856, 438 + 438 = 0876, 459 + 459 = 0918, 469 + 469 = 0938, 479 + 479 = 0958, 734 + 734 = 1468, 765 + 765 = 1530, 836 + 836 = 1672, 846 + 846 = 1692, 867 + 867 = 1734, 928 + 928 = 1856, 938 + 938 = 1876.
Using the expressions T = n, W = x, O = y, F = c, U = f, R = a, in TWO + TWO = FOUR, simply remember that for the answer to be right, you must have a number where: y = 2n or y = 2n + 1 if x = 5<.
Thank you for reading my solution, I hope it is helpful!
Edward McAllister