Saudi Arabia booklet, 2024

The answer is NO.
Consider all words of length 13 and continue them periodically in both directions to create the sequences. For details, take $x_1x_2\dots x_{13}$ as some original word of length 13 then add $x_1, x_2, x_3, \dots$ to the right and add $x_{13}, x_{12}, x_{11}, \dots$ to the left will result:
$$\dots x_{11}x_{12}x_{13}(x_1 x_2 \dots x_{13})x_1 x_2 x_3 \dots$$
There are two identical original words as $AAA\dots A$ and $BBB\dots B$ will generate the unique sequences. For the other words, each of them will generate the same sequences with another 12 words as follows:
$$(x_2x_3x_4\dots x_{13}x_1),\ (x_3x_4x_5\dots x_{13}x_1x_2),\ \dots,\ (x_{13}x_1x_2\dots x_{12})$$
So in total, the number of distinct sequences is
$$\frac{2^{13}-2}{13}+2>600.$$
By the definition of the sequences, one can check that any word of length at least 13 may appear only in 1 sequence. Each word of length $n < 13$ may appear in at most $2^{13-n}$ sequences, since we fix $n$ letters and choose another $13-n$ other letters with $2^{13-n}$ ways to form a word of length 13 (some of them may coincide). And note that
$$(2^8 + 2^7 + \dots + 2^0) + 17 = 2^9 + 16 < 600$$
so we will have available sequences.