I think there are 2 main sources of confusion:
First, because of gravity, extending your arms feels like work. We're only interested in the radial movement, though, and in this direction, the skater's arms are pulled by the centrifugal force (in the long tradition of spherical cows in vacuum, we could replace the figure skater with two beads on a spinning rod).
Second, the idea of rotational energy as kinetic energy. The relevant work variable is (as already mentioned) the radial extension of the skater's arms, and as far as that's concerned, rotational energy plays the part of potential energy.
Think of the skater pulling in her arms as compressing a spring, and extending the arms as its release.
Going by either the bead or spring model, the rotational energy gets converted into kinetic energy of the arms, accelerated by the centrifugal force in direction of the radial work variable and ultimately dissipating via vibrations when the arms abruptly reach maximal extension.
Of course, if the skater doesn't let her arms be accelerated and slowly extends them instead, the energy dissipates right away, which might be the more realistic approach.
Here's an example of why this is done using a gravitational context.
Kinetic energy
The kinetic energy of a system of particles is defined to be $K_\text{tot} = \sum_i K_i$, where the sum runs over all particles in the system. This is just a definition, and choosing to use such a definition allows one to speak of the kinetic energy $K_i$ of the $i$th particle, for example, since it's just one of the terms in a sum.
Okay, that was boring.
Potential energy
Gravitational potential energy $U_\text{g} = -\frac{Gm_im_j}{d_{i,j}}$ is defined for pairs of particles.$^\text{1}$ Mathematically you can see that in the subscripts of the masses $m_i$ and the distance between the two particles $d_{i,j}$. Why? Because to determine potential energy from scratch, you must consider how two objects interact, which in classical mechanics is where forces come in. And forces themselves come in pairs.
Now, the fact that they come in pairs means one must be careful when determining the potential energy in some given situation, otherwise you might over-count. Suppose, for example, there are 4 particles with masses $m_i$, and the distance between the $i$th and $j$th objects is $d_\text{i,j}$. What, then, is the potential energy of the masses and distances are all given? The answer is
$$U_\text{g}=-G\left( \frac{m_1m_2}{d_{1,2}} + \frac{m_1m_3}{d_{1,3}} + \frac{m_1m_4}{d_{1,4}} + \frac{m_2m_3}{d_{2,3}} + \frac{m_2m_4}{d_{2,4}} + \frac{m_3m_4}{d_{3,4}} \right)$$
First, note that since there are six terms you can't make a one-to-one associating of particles with potential energy. Second, if someone tried to associate potential energy with a single particle, then that person might mistakenly include too many terms in their sum:
$$U_\text{g}\ne-G\left( \underbrace{\frac{m_1m_2}{d_{1,2}}}_\text{a term for particle 1} + \underbrace{\frac{m_2m_1}{d_{2,1}}}_\text{a term for particle 2} + \cdots \right)$$
This incorrect expression would over-count the number of terms in $U_\text{g}$.
To avoid such over-counting, and to make our language more inline with the mathematics, we tend to speak of the potential energy of the system.
$^\text{1}$The expression $-\frac{Gm_1m_2}{d_{1,2}}$ is a more general form of the more-common $U=mgh$, but that's a separate issue.
Best Answer
I agree with CuriousOne that the example is more confusing than helpful, but this is the way I would explain it.
Suppose you take a spring, place it on the ground then compress it. If you now suddenly let go of the spring it will rebound and bounce upwards off the ground:
The spring clearly has work done on it because its kinetic energy increases and that increase must have come from somewhere. However the ground can't have done any work on the spring because the ground hasn't moved. It should be obvious that the potential energy in the compressed spring has been converted into kinetic energy of the uncompressed spring - in effect the spring has done work on itself. This is what your book means by:
i.e. potential energy of the compressed spring has been converted into kinetic energy of the uncompressed spring.
In the case of the skater the skater's arms correspond to the spring and the rail corresponds to the ground. The skater's arm isn't a spring, of course, because it's chemical energy not potential energy being converted to kinetic energy by the skater's muscles. Nevertheless the same principle applies.