The record for the most mutual checks is known, but what about the record for double checks? There are a few rules though so this question can actually have an answer.
A checkmate move only counts it is a double check.
Threefold repetition is in effect to avoid infinite loops. The count stops right away if it happens, although the last move still counts.
The position must be legal.
I have four catergories that I have come up with. Promoted pieces allowed and not allowed are two criteria. The next two are “forced” and “unforced.” The first means that Black must allow themselves to be double-checked, and in the second they play along.
I hav done some initial work myself. Who can find more?
No Promoted Pieces, Unforced-6 Double Checks
[FEN "4K3/8/8/3p4/k1p1N3/1p4N1/P5BB/RRQ5 w - - 0 1"] 1. axb3+ Kb5 2. bxc4+ Kc6 3. cxd5+ Kxd5 4. Nc3+ Kd6 5. Nf5+ Kc5 6. Na4#
No Promoted Pieces, Forced-5 Double Checks
[FEN "kb1r4/P1p5/1P2P3/6NB/7B/8/8/RRQ4K w - - 0 1"] 1. axb8=Q+ Kxb8 2. bxc7+ Kc8 3. cxd8=Q+ Kxd8 4. Nf7+ Ke8 5. Nd6+ Kf8
Promoted Pieces, Unforced-9 Double Checks
[FEN "K5b1/5p2/4p3/3pR3/k1p1N1N1/1pQ5/P1B5/RR1R1RRR w - - 0 1"] 1. axb3+ Kb5 2. bxc4+ Kc6 3. cxd5+ Kd7 4. dxe6+ Ke8 5. exf7+ Kf8 6. fxg8=Q+ Kxg8 7. Nh6+ Kh7 8. Nf6+ Kh8 9. Nf7#
Promoted Pieces, Forced-8 Double Checks
[FEN "K4Bb1/5p2/4p3/3p4/k1p3N1/1pR5/P1B5/RQ1RRRRR w - - 0 1"] 1. axb3+ Kb5 2. bxc4+ Kc6 3. cxd5+ Kd7 4. dxe6+ Ke8 5. exf7+ Kxf8 6. fxg8=Q+ Kxg8 7. Nh6+ Kh8 8. Nf7#
As Hauke Reddmann noted in a comment, Ba1 Bb1 Rb2-Kc3 is a well-known double checking mechanism, but no origin is given. The earliest instance I have found, and the likely origin of it, is from 1919, which is a century ago as @Evargalo proposed a comment.
[Title "Alain Campbell White, Pittsburgh Gazette Times 4/1916 #12"] [FEN "2q5/2pp4/3pr3/4pb2/K1p2pn1/2bn1kp1/3pr1R1/6BB w - - 0 1"] 1. Rf2+ Ke3 2. Rf3+ Ke4 3. Re3+ Kd4 4. Re4+ Kd55. Rd4+ Kc5 6. Rd5+ Kc6 7. Rc5+ Kb6 8. Rc6+ Kb7 9. Rb6+ Ka7 10. Rb7+ Ka8 11. Ra7+ Kb8 12. Ra8#
It turns out that the record for the double checks in a checkmate problem is 13 in the mainline. It cannot contend for the forced record though since Black has other options.
[Title "Стојнић, Драган Бабић, Миломир, The Problemist 2004-05, #13"] [FEN "3q1nKB/R1P1kPRB/N3p3/1p1n2p1/2r2p2/1p2b3/P2pb2N/3r4 w - - 0 1"] 1. c8=N+ Kf6 2. Rg6+ Kf5 3. Rf6+ Ke5 4. Rf5+ Ke45. Re5+ Kd4 6. Re4+ Kd3 7. Rd4+ Kc3 8. Rd3+ Kc2 9. Rc3+ Kb2 10. Rc2+ Kb1 11. Rb2+ Ka1 12. Rb1+ Kxa2 13. Nb4#