Edward and Thomas from Dartford Grammar School worked out that:

Starting at zero (with both ropes parallel), the sequencetwist, twist, twist, turn , twist, twist, twist, turn , twist, twist, twist, turn

takes us to:$$0, 1, 2, 3, -\frac{1}{3}, \frac{2}{3}, \frac{5}{3}, \frac{8}{3}, -\frac{3}{8}, \frac{5}{8}, \frac{13}{8}, \frac{21}{8}, -\frac{8}{21}$$

Oliver from Olchfa School also worked this
out and noted that:

If the 'moves' "twist (x3), turn" are repeated, the fractions
produced only include numbers from the Fibonacci sequence in the
numerators and denominators

Terence from The Garden International School in Kuala Lumpur, William from Shebbear College and Akshita from Tiffin Girls' School worked out how to disentangle themselves:

The following sequence takes us back to zero:

twist, turn , twist, twist, turn , twist, twist, twist, turn , twist, twist, twist, turn , twist, twist: $$ \frac{13}{21}, -\frac{21}{13}, -\frac{8}{13}, \frac{5}{13}, -\frac{13}{5}, -\frac{8}{5}, -\frac{3}{5}, \frac{2}{5}, -\frac{5}{2}, -\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}, -2, -1, 0$$

Both sequences were written up by Sakib from
Swanlea Secondary School - you can find his work here .

Oliver from Olchfa School worked this out
and added:

To distangle this, we must reduce the fraction to the form
1/n.

We keep twisting the negative fraction until we get the first
positive fraction (< 1) which is then turned.The procedure is repeated until we get to 1/n.

Now we turn again, then twist n times to get the fraction back to 0.

The full proof is given in the problem More Twisting and Turning