This sounds complicated at first since there are so many ways in which a certain final matchscore can come about. Then I remembered that it does not matter when you get your luck, only that you get it. So my guess is that my chance of winning a 5 point match with a final score of 5-2 should be the same as winning it from 0-2.  If this is correct, then the problem reduces to looking up the probability for 5-away/3-away in a MET (which is where the assumption of equal playing strengths is needed). The general case would amount to looking up n-away / (n-m)-away.

I don't feel 100% confident yet about this answer, please correct if I have overlooked something.

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Correction:
My answer cannot be right since winning from 0-2 can result in other final scores than 5-2 and the probabilities for 5-3 and 5-4 are included in P_MET(5-away / 3-away).

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Correction 2:
Since P_MET(5-away / 3-away) gives my winning probability for the case of my opponent getting at least 2 points, deducting the probability of him getting at least 3 points should give the correct answer: P(I win 5-2) = P_MET(5-away / 3-away) - P_MET(5-away / 2-away).
For the general case we get: P(I win n-m) = P_MET(n-away / (n-m)-away) - P_MET(n-away / (n-m-1)-away)

Some numbers (using Neil Kazaross's MET KazarossXG2.xml bundled with gnubg):

Probabilities are given as P(the match ends 5-n).

P(5-0) = 15.4%
P(5-1) = 14.0%
P(5-2) = 19.2%
P(5-3) = 19.6%
P(5-4) = 31.6%

At first I thought this looks plausible: The players have equal strength, so close match scores should occur more often than more extreme ones. But can 5-4 really be expected to occur that much more often than 5-3? After all, at 3-3, cubing is mandatory. So I let gnubg play 15 5-pointers against itself, both players at world class strength. Result:

5-0   5 times
5-1   0 times
5-2   7 times
5-3   2 times
5-4   1 time

15 matches is not much, but still it seems there is something seriously wrong with my theory, back to the drawing board...

I can't make a meaningful contribution to the mathematical argument, but I can count. Of the 258 5 point matches in my database, the breakdown of results is as follows.
5-0 103
5-1 33
5-2 49
5-3 35
5-4 38.
All the matches involve me of course against a wide range of humans. Is equal skill important to the answer?

Pck's answer, P(I win n-m) = 2 * [P_MET(n-away / (n-m)-away) - P_MET(n-away / (n-m-1)-away)], fulfills two requirements: it uses the MET, and the n probabilities are positive and add up to 1. However, I am not convinced for three reasons:

I'm pretty sure now that my answer is not only wrong but off by a mile. The reasons you gave, plus dorbel's and my own numerical data make that clear. The confusion is this: I based everthing on the idea that with regard to winning, it is immaterial when you get your luck in a match. It is only important that you get a certain total amount. That is certainly correct, but it pertains to winning the match, not to the question of what the final score will be. The final score will indeed depend on when you get your luck. You can be unlucky first, then very lucky and win 5-3, or you can be moderately lucky from start to finish and win 5-0. In both cases you win with the same amount of luck (in fact, and rather trivially, all matches are won with the same amount of luck given a certain fixed difference of skill of the players). I think this is compatible with what the third point in your list says (red probabilities not cancelling out).

Therefore, "P_MET(5-away / 3-away) gives my winning probability for the case of my opponent getting at least 2 points" is wrong and the entire thing collapses.

If the probabilities for each k_W and k_L for each position in the MET were known, the problem can easily be solved using a Markov chain. Unfortunately, I don't know them, but their information should somehow be incorporated in the MET.  

That was my initial thought too, and I could not come up with a quick way to calculate P(k_W) and P(k_L) either. Obviously they depend on the match score. Perhaps it is possible to derive them recursively in a manner similar to the way METs are created? I haven't thought this through in any way yet, but the probability of the cube being turned may be a big problem.