Amortized analysis for disjoint sets' find-set(x) function (from CLRS)
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 $0 \le \operatorname{level}(z) &lt; \alpha(n)$. 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. 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 $\operatorname{level}(x) = \operatorname{level}(y)$. 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.