Keep trying whatever you can, then whenever you have exhausted those options, move on to the next thing to try.  Keep in mind, this is our instructions:

Do this:

Let's decrypt this:

"A MINGDIOTHXIM IODPINGUWAM TA
ING INGTHIOWSH IC AEPAEGO MIA EO TW
O FYUTJ THJEAECNEA CEAE R
OUX DWHBINGFACJU OJIOEAILGH BU
J XACWIO MOPYDIOLP EOITHR YGM X
OEGS MPJOED OETIOOE TINGI MINGYAAE ING R
YJ NLUXB HPJOEA GAEW XSDC UT 
OEHEONPT EAYBUG GAEO NGOELMBTH Y
IOOEEOY MCOETH ELREOPRDY XD INGYP 
SEXW CGFDRE DEWOEOOE HOOE BS H
EAACIMXJME ML GOTTHXEEARGY

ING AEMIW SB XM SEOGDTHAE FIOABCU WR B
EAFXIMUCPHINGLEO MSDIOPF BHEOC 
EOE EAEIGXGS ESEOTHTHES THOE CEAIM
CX DJU BEI"
The last currently unsolved section of the LP (pointed at, perhaps, by the most recent hint) may have given up a clue. The "diffs" distribution, but which we mean D(n) = (C(n) - C(n-1)) % 29, seems to have properties indicative of a periodic polyalphabetic substitution. The thumbnail of this post is evidence to that effect, with a well-motivated guess about where possible interrupters could be. What you see is a Friedman-style IoC analysis for periods of 1 to 50, using the interrupters in the subsequent discussion.

Note the large spikes at 13, 26, and 39. 
Image
igloo party
OP
 — 11/01/2023 11:03 AM
Steps to reproduce/thought process involved:

1) Assume based on how well ciphertext autokey explains the doublet count that a similar algorithm is at play.

2) Assume that the cipher takes the form C(n) = C(n-1) + f(P(n),K(n)), for an invertible (given K(n) and C(n) we can find P(n) uniquely) function f, cipher text at position n, C(n), and plaintext P(n), and keystream K(n).

3) Further assume that at least on some pages it is possible that K(n) is periodic.

4) To investigate f directly, we need to look at D(n) = (C(n) - C(n-1)) (mod 29) = f(P(n),K(n)).

5) Compute the index of coincidence of D(n) for each page with different possible key periods (1 to 20 in my case).

5) Notice that on page 54 for period = 13 we get 1.3344 and on page 55 for the same period = 13 we get 1.8590. Noticing that these are both under the same red rune section compute for the whole section. We get 1.3792, which is quite impossible (about 0.05% chance) for 300ish runes.

6) This isn't high enough to be natural language, but we know there are probably interrupters. We have two options: A) try to decrypt a short section (hard if we can't guess the form of the unknown function f) or B) try to guess the locations of interrupters so we have more data to work with.

7) Taking option 2, we make the guess that interrupters all have the same value, not in the ciphertext, but in D. The search space is small enough that we can brute force any combination of interrupters for each value 0-28.

8) As a result we get that for interrupters at 62    73   109   168   180   306 we get an IoC of 1.7201. All of these locations are where D(n) = 1. Other interrupter sets get similar

9) Next step is attempt to solve the periodic substitution or brainstorm some good guesses to lower the search space.

It is worth noting, the author has a solver that seems to work on texts of a comparable size if the function f were an affine cipher for each slice and this has failed to find a solution.
Other suggested future work includes investigating interrupters that interrupt in BOTH layers of the proposed encryption.

Both analysis and a general attempt at solving using the work thus far are both underway, at least by OP, but hopefully perhaps others as well.
igloo party
OP
 — 02/09/2024 9:27 AM
It has been very understandably pointed out that for various interrupter sets the diffs sequence which this post studies has, at a glance, awkward and disparate period 13 slice IoCs. However, further analysis seems to suggest that this gut instinct does not seem to be backed up by data:

The data: 1e4 rune streams or the appropriate length each with a 1 gram IOC which is sampled from the true distribution of runeglish encrypted with a random quagmire 4 cipher (of period 13 of courses). We show a histogram of each the mean, standard deviation, max, min, and range, of the random data with the observed deltas with the interrupters 6    62    74   102   169   193 marked as a red vertical line. It is clear that the observation is close to the mean of each distribution.