I assume that by $w$ you mean $\omega$.
Then showing $\omega\omega=(\omega+\omega)\omega$ is relatively easy.
The first one is the set $\omega\times\omega$, the second one is $(\omega\times\{0\}\cup(\omega\times\{1\}))\times\omega$, both of them with antilexicographic order, i.e. the second coordinate is more important.
The isomorphism between the two sets is $(a,2b)\mapsto((a,0),b)$ and $(a,2b+1)\mapsto((a,1),b)$.
In fact, this can be visualized quite easily $\alpha\times\beta$ means that you take ordinal $\beta$ and replace each element by a copy of $\alpha$. I.e. $\alpha\beta$ consists of "$\beta$-many" copies of $\alpha$ which you put one after each another, and they are ordered according to $\beta$.
In this sense $\omega\omega$ consists of $\omega$ copies of $\omega$.
What is $(\omega+\omega)\omega$? I've put several pairs of copies of $\omega$ after each other. This is the same as $2\omega$ copies of $\omega$, but $2\omega=\omega$. (If I replace each element in $\omega$ by a pair of elements, then I do not change the order type.)
At the moment I do not have time to draw a nice picture illustrating this but this might help too:
- $\omega$ looks like $\ast \ast \ast \cdots$
- $2$ is simply the two points $(\circ \circ)$
- To get $2\omega$ we simply replace each $\ast$ by $(\circ \circ)$ and we get $(\circ \circ) (\circ \circ) (\circ \circ) \cdots$
- Of course we can take parentheses away (they are just an auxiliary thing to show where the copies of $2$ were added), so this is simply $\circ \circ \circ \circ \circ \circ \cdots$
- The two things look the same, only we have used $\ast$ to denote the elements in one of them and $\circ$ in another one.
You can prove the same thing using some rules for cardinal arithmetic, namely you have $\omega+\omega=\omega\cdot 2$ so it suffices to show that $2\omega=\omega$ (which might be easier for you). If you know both these things you get $(\omega+\omega)\omega=(\omega2)\omega=\omega(2\omega)=\omega\omega.$
More importantly I would like to ask you to clarify the second part (perhaps ti would be even better to post it as a separate question).
You have something like this:
Let $A$ and $B$ be sets. Let $A$ be ordered by $\le_G$ and $B$ by $\le_H$. Let $f$ be an isomorphism such that $x\le_G y$ implies $f(x)\le_H f(y)$. Now, reorder $A$ by an order relation $\le_{G'}$. Then does there exist an isomorphism $f'$ such that $x\le_{G'}y$ implies $f'(x)\le_H f'(y)$?
What precisely do you mean under isomorphism? If you mean isomorphism of ordered sets $(A,\le_G)$ and $(B,\le_H)$, then the definition of isomorphism includes that $x\le_G y$ $\Leftrightarrow$ $f(x)\le_H f(y)$. So your sentence such that $x\le_G y$ implies $f(x)\le_H f(y)$ seems to be redundant. Now if there are no conditions on $\le_{G'}$, you are basically asking whether two orders of the same set must be isomorphic. This is certainly not true. (But maybe I've misunderstood the question.)