Proof that cube has 24 rotational symmetries
The point of the comment in the book is this: given a cube, pick one corner and call it $A$. You can certainly rotate the cube so that $A$ either stays put or moves to any chosen other corner, of which there are 7. So you have 8 choices for where you want $A$ to be.
Once you've done this, the other corners are obviously not free to move as they wish. Let $B$ be some corner adjacent to $A$, meaning it shares an edge of the cube with $A$. If A moves to $A'$, $B$ must move to some corner adjacent to $A'$ - otherwise the edge $AB$ gets distorted. How many corners are adjacent to $A'$? That's where the 3 comes from.
Finally you'll need to convince yourself that once $A$ and $B$ have been placed, all other corners are "forced" - their final location is already fixed. See if you can do that just using the relationships of adjacency, sharing a face etc.
A cube has 12 edges, so it has 24 oriented edges (each edge can be oriented in exactly two ways) It is pretty obvious, if you have a cube to play with, that the group $G$ of rotations acts simply transitively on these orientied edges. Therefore $G$ has 24 elements.
Exactly the same argument counts the number of rotational symmetries of each regular polyhedron, of course.
Counting the possible orientations of the cube we know that there are 24 rotational symmetries, by considering faces, edges or corners and the their respective number of orientations, giving 6 * 4 = 12 * 2 = 8 * 3 = 24.
It seems you also want to explicitly know the rotations, instead of just counting the rotational symmetries. The rotations are:
1 identity rotation that does nothing.
9 rotations around axes through the middle of one of 3 pairs of opposing faces, with rotations of 90°, 180° and 270°.
6 rotations around axes through the middle of one of 6 pairs of opposing edges, with a rotation of 180°.
8 rotations around axes through one of 4 pairs of opposing corners, with rotations of 120° and 240°.