Rachel thought through the problem like this:

For fractions $-\frac{n}{2}$

twist until reaching $\frac{1}{2}$, then turn to get $-2$, and twist twice.

For fractions $\frac{2}{n}$, start by turning once, then repeat as above.

This also works for $3$ instead of $2$, etc. and I think can be generalised:

for $-\frac{n}{m}$: twist to $\frac{1}{m}$, turn to -m, twist to $0$

and for $\frac{n}{m}$: turn to $-\frac{m}{n}$, twist to $\frac{1}{n}$, turn to -n, twist to $0$

I think that this shows that any fraction can revert to $0$.

Rachel's initial thinking regarding negative halves is absolutely correct.
Her follow-up argument is along the right lines:
if the fraction is negative, then twist (+1)
if the fraction is positive, then turn (-1/x)
However, it sometimes requires a longer sequence than the one that Rachel suggests.
You cannot guarantee that you will always arrive at a unit fraction (numerator of 1) after a sequence of twists:

e.g. starting at $-\frac{21}{13}$, two twists take us to$-\frac{8}{13}$ and $\frac{5}{13}$, not $\frac{1}{13}$
Oliver from Olchfa School sent us thiscomprehensive solution. Thank you Oliver.