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
$\frac{1}{13}$