$\omega + 1$ 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.
No, there are many more ordinals -- you've only seen the tip of the iceberg! The first ordinal not on your list is usually called $\omega \cdot 2$ (read "omega times two"), and it consists exactly of the natural numbers, the set $\omega$, and all of $\omega$'s iterated successors. That is,
$$
\omega \cdot 2 = \{0, 1, 2, \ldots, \omega, \omega^+, (\omega^+)^+, \ldots\}.
$$
(Usually what you call $\omega^+$ is called $\omega + 1$, and $(\omega^+)^+ = \omega + 2$, and so on.) You could call $\omega \cdot 2$ cat if you want to, but $\omega \cdot 2$ is definitely a more common name for it. :-)
Presumably you can now guess at what $\omega \cdot 3$ is, and $\omega \cdot 4$, up to $\omega \cdot n$ for any natural $n$. But we're not done of course, after that you can add up all of those (and their successors) and get $\omega \cdot \omega$, or $\omega^2$. And then you might be able to guess what $\omega^2 \cdot 2$ is, and $\omega^2 \cdot 3$, and maybe $\omega^2 \cdot \omega$ -- and then the notation $\omega^3$ starts to make sense. And then $\omega^4$, and then eventually $\omega^\omega$. And still this is only a very small tip of the iceberg!
Enjoy exploring!
Edit: based on the clarification in the comments, maybe this better answers your question.
Proposition: For any ordinals $\alpha, \beta$, either $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$.
Proof: Suppose towards a contradiction that this is false; that is, $\alpha$ contains elements not in $\beta$, and vice versa. Since $\alpha$ is a well-order, we can find the least element of $\alpha$ that is not in $\beta$; call it $\alpha'$. Similarly, let $\beta'$ be the least element in $\beta$ that is not in $\alpha$.
Let us show that $\alpha' = \beta'$, contradicting our assumption. It suffices to prove $\alpha' \subseteq \beta'$ and $\beta' \subseteq \alpha'$ -- WLOG let's prove the former.
Take $\alpha'' \in \alpha'$. This means that $\alpha'' \in \alpha$, and in the ordering in $\alpha$, $\alpha'' < \alpha'$, so by definition of $\alpha'$ we have that $\alpha'' \in \beta$. We cannot have that $\beta' < \alpha''$, because that would mean $\beta' \in \alpha''$ and thus $\beta' \in \alpha$. Similarly, we cannot have $\beta' = \alpha''$, because then we would also have $\beta' \in \alpha$. Thus we have $\alpha'' < \beta'$, or in other words $\alpha'' \in \beta'$. It follows that $\alpha' \subseteq \beta'$.
This is a contradiction, so $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$. $\square$
That tells you, basically, that all the ordinals there are, are in one long line. Every ordinal (as a set) consists exactly of all ordinals below it, and there is no "branching". In particular, the natural numbers are the only finite ordinals, and every infinite ordinal contains the natural numbers.
Best Answer
Kenneth Kunen's Set Theory: an Introduction to Independence Proofs develops the Von Neumann ordinals pretty much right out of the gate.