I suspect that the definition of $\omega^2$ was intended to read
$\omega^2=\bigcup\{\omega\cdot n:n<\omega\}\;,$
where $\cdot$ indicates ordinal multiplication. Using $\longrightarrow$ to represent a set ordered in type $\omega$, we can picture $\omega\cdot n$ as
$\underbrace{\longrightarrow\longrightarrow\ldots\longrightarrow}_{n\text{ arrows}}$
and get the following picture:
$\begin{array}{rl} \omega\cdot 0:\\ \omega\cdot 1:&\longrightarrow\\ \omega\cdot 2:&\longrightarrow\longrightarrow\\ \omega\cdot 3:&\longrightarrow\longrightarrow\longrightarrow\\ \vdots\quad&\qquad\vdots\\ \omega^2=\omega\cdot\omega:&\underbrace{\longrightarrow\longrightarrow\longrightarrow\longrightarrow\longrightarrow\ldots}_{\omega\text{ arrows}} \end{array}$
The order type $\eta+1$ is what you get when you add a right endpoint to $\Bbb Q$; it’s the order type of $\Bbb Q\cap(0,1]$, for instance. The order type $(\eta+1)\cdot(\eta+1)$ can then be visualized as follows.
Let $A=\Bbb Q\cap(0,1]$, and let $X=A\times A$. Let $\le$ be the usual linear order on $A$, and let $\preceq$ be the reverse lexicographic order on $X$ defined as follows: if $\langle p,q\rangle,\langle a,b\rangle\in X$, $\langle p,q\rangle\preceq\langle a,b\rangle$ iff either $q, or $q=b$ and $p\le a$.
$X$ is the set of points of $(0,1]\times(0,1]$ with rational coordinates. Speaking pictorially, if $x$ and $y$ are points of $X$ with $x$ to the left of $y$, then $x\preceq y$; if $x$ and $y$ are on the same vertical line, then $x\prec y$ iff $x$ is below $y$.