It sounds like they're describing a freely rotating platform. If the platform and girl start at rest, and if there's no frictional torque on the platform from its axle, then Conservation of Angular Momentum tells us that if the girl starts walking clockwise around the platform, the platform itself must move counterclockwise so that the total angular momentum is still 0.

The platform's angular momentum must be of the same magnitude as the girl's angular momentum, because the girl and platform's angular momenta have to cancel out.

Angular momentum can be calculated by L = I*w. We can treat the girl, who moves in a circle around the axis of rotation that passes through the platform's axle, as essentially a point mass moving in a circle. Thus I_girl = m*r^2. So when the radius from the axle to the girl goes down by a factor of 1/2, her angular inertia goes down by 1/4. Her angular velocity is still the same, so her angular momentum has also dropped by 1/4. This means the angular momentum of the platform must also be 1/4. The angular inertia of the platform doesn't change, so the angular velocity of the platform is also 1/4.

Another way to look at this is to calculate angular momentum by L = v*r*sin(theta). Theta is 90 degrees, so we can ignore that part. When the girl moves to a distance of 1/2 her initial distance, as you point out, her velocity is 1/2 what it used to be. L thus goes down by a factor of 1/2 because of her decreased v, and another factor of 1/2 because of her decreased r. The total decrease in L is thus 1/4. Again, this means the angular velocity of the platform must drop by 1/4 for Angular Momentum to be Conserved.