Unfortunately, I am not a latex/markdown guru, either. Let's see if we'll be able to make things work nonetheless.
The idea is actually quite simple. Just like in the book, let us assume that the find path consists of s nodes. We disregard the root (because its potential doesn't change at all after the operation) and the node we upon which find_set is called (because it may have rank 0, which means its potential might not change). This leaves us with s-2 nodes.
Now, consider any node z among the remaining s-2 nodes. As explained on pages 576-577 (I am using the 3rd edition), we have 0 <= level(z) < alpha(n). In other words, the level of z must be between 0 and alpha(n) - 1, inclusive. Indeed, it is the fact that we have a finite number of options for the levels of the aforementioned s-2 nodes that will help us prove the assertion presented by the authors.
They say one example is worth a thousand words, so let's do one. 7
Suppose we have 8 nodes between the root and the one upon which find_set was called. Suppose also that alpha(n) turns out to be 4. Here's one possibility for the levels of the nodes, given in the order they appear in the find path:
2, 0, 1, 3, 2, 3, 1, 0
The book claims that there are at most alpha(n) (4 in our case) nodes x for which there doesn't exist a node y later in the path such that level(x) = level(y). Take a good look at the sequence; you'll see that the claim holds for our example.
Now, the million dollar question. Why? Think about the last alpha(n) nodes. We could assign different levels to each one (as I have done for this example) so that none of them have a complementary node higher up the find path with the same level. However, that is the most we can have; every other node will now have some ancestor with the same level. Long story short, no matter how hard we try, we cannot have more than alpha(n) nodes without an ancestor of equal level; such a scenario would require more options for the level of a node than are available.
Put it all together, and it is evident that there exists another node with the same level down the line for at least s - alpha(n) - 2 of the s nodes, just like the book claims.
As for why the authors have found it unnecessary to provide an explanation for their claim, you are asking the wrong guy. 743