Edward and Thomas from Dartford Grammar School worked out that:

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