$\omega + 1$ is a good example. $\omega + 1$ is the ordinal $\{0, 1, 2, \ldots, \omega\}$; as an ordering, think of it as $\omega$ with one more element at the end. It is greater than $\omega$, because it has $\omega$ as a proper initial segment; for ordinals, that's what "greater" means. But it's not bigger than $\omega$ - there's a bijection between $\omega$ and $\omega + 1$, given by the function $f$ which takes $0$ to $\omega$ and $n$ to $n - 1$ for all $n > 0$.
By contrast, the ordinal $\omega_1$, defined as the first uncountable ordinal, is bigger than all of its predecessors, by definition - if $\alpha < \omega_1$, then $\alpha$ can't be uncountable, so there is an injection from $\alpha$ to $\omega$. But there's no injection from $\omega_1$ to $\omega$, by definition, because that would make $\omega_1$ countable. So $\omega_1$ is a cardinal, often denoted $\aleph_1$.
The key idea here is that the author is using "bigger" to refer to size, not ordering - that is, "bigger" is a statement about whether a certain injection exists, not where an ordinal appears in the standard ordering.
On the other hand, you asked about an ordinal that is "equal to or smaller than some of its predecessors". This can't happen. An ordinal is never equal to its predecessors, because different ordinals are always different - it's like asking whether there's a number that's equal to a different number. And since every ordinal has an injection into all of its successors, ordinals can't decrease in size. The only thing that can happen is the example I've outlined above, where we have an ordinal that's the same size as its predecessor. Note that this means that, for ordinals, "the same size as" and "equal to" do not mean the same thing.
For each proper initial segment $C$ of $B$ we have a unique order isomorphism between $C$ and an ordinal $\gamma \in \kappa$.
This gives us a unique order isomorphism $f_C : A \cup C \to \alpha + \gamma$, and by uniqueness all of these isomorphisms agree on their common domain.
So we can take the union $f = \bigcup f_C$, which is an order isomorphism between $A \cup B$ and $\bigcup_{\gamma \in \kappa}(\alpha + \gamma)$.
Best Answer
It is not the case for any ordinal.
Consider the ordinal $\omega + 1 = s(\omega) = \mathbb{N} \cup \{\mathbb{N}\}$. Then $\omega+1$ and $\omega = \mathbb{N}$ are in bijection with each other by the bijection
$f(\mathbb{N}) = 0$
$f(n) = s(n)$ for all $n \in \mathbb{N}$
But clearly $\omega$ precedes $\omega + 1$.