Let $A_n=\{0,1,\ldots,n-1\}\subseteq\mathbb{N},n>0$, so that $|A_n|=n.$ Order on this set is the usual order $<$ on the naturals. For this example, let's use $A_2=\{0,1\}$, $\mathbb{N}$ and lexicographical ordering on the set $A_2\times\mathbb{N}$. Since lexicographical order considers the first element, the element of $\{0,1\}$, first, we can arrange the elements as $(0,0),(0,1),(0,2),\ldots,(1,0),(1,1),(1,2),\ldots$ giving us the order type $\omega+\omega=\omega\cdot 2.$
If we instead look at the set $\mathbb{N}\times A_2,$ we get the order $(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),\ldots$ giving us the order type $\omega.$
Question 1: If I want to construct a set with order type $\omega\cdot n$, is this process always sufficient, that is, is $\operatorname{ord}(A_n\times\mathbb{N})=\omega\cdot n$ for all $n$?
Question 2: Is the particular reason the notation for $\omega+\omega$ is specifically $\omega\cdot 2$ in the way the multiplication is defined inductively on the ordinal on the right side? Any historical/practical reason for this?
Looking at the order the sets appear in the Cartesian product would intuitively give $2\cdot\omega,$ similarly to cardinality of Cartesian product of finite sets.