Why is $\omega+1+\omega+1 = \omega+\omega+1$? Why isn't it $\omega+\omega+2$?
On the other hand, is it true that $\omega+\omega = 2\omega$?
Also why is it that $2 \omega=\omega$?
Here $\omega$ is taken to be the limit ordinal which is just $\mathbb{N}$.
I am really confused as to why ordinals can't add/multiply just like natural numbers do, since they are essentially the same thing?
 A: 
I am really confused as to why ordinals can't add/multiply just like natural numbers do, since they are essentially the same thing?

The ordinals generalize the natural numbers in a certain sense, but that does not mean that every property of the natural numbers carries over to the ordinals. Weird though it may be, neither ordinal addition nor ordinal multiplication are commutative.
Ordinals are weird beasts. It's best at first to think of an ordinal as a particular linear order - a collection of dots laid out in a line (not every linear order corresponds to an ordinal, of course, but every ordinal corresponds to a linear order). "$\alpha+\beta$" is what you get by putting a copy of $\beta$ after a copy of $\alpha$; "$\alpha\cdot \beta$" is what you get when you replace each point in $\beta$ with a copy of $\alpha$.
Reasoning pictorially, $1+\omega$ is $$1+\omega\quad=\quad{\large\bullet}\quad +\quad{\large\bullet}+{\large\bullet}+{\large\bullet}+{\large\bullet}+...\quad=\quad{\large\bullet}+{\large\bullet}+{\large\bullet}+{\large\bullet}+...\quad=\quad\omega,$$ while $\omega+1$ is $$\omega+1\quad=\quad {\large\bullet}+{\large\bullet}+{\large\bullet}+{\large\bullet}+...\quad+{{\large\bullet}}\quad=\quad {\large\bullet}+{\large\bullet}+{\large\bullet}+...{\color{red}+ \color{red}{\large\bullet}},$$
 and this latter does not look like $\omega$ (unlike $\omega$, it has a last element).
Here's another picture-argument: first, thinking about $2\omega$, we have $$2\omega\quad=\quad (2)+(2)+(2)+...\quad=\quad ({\large\bullet}+{\large\bullet})+({\large\bullet}+{\large\bullet})+...\quad=\quad {\large\bullet}+{\large\bullet}+{\large\bullet}+{\large\bullet}+...\quad=\quad\omega,$$ but $\omega2$ is $$\omega2\quad=\quad(\omega)+(\omega)={\large\bullet}+{\large\bullet}+{\large\bullet}+...{\color{\red}+ \color{red}{\large\bullet}}+{\large\bullet}+{\large\bullet}+...,$$
 and this does not look like $\omega$ (unlike $\omega$, it has an element with no immediate predecessor).

In your title problem, since commutativity fails we can't argue that $$\omega+1+\omega+1=\omega+\omega+1+1=\omega+\omega+2;$$ the right answer is instead to see that $$\omega+1+\omega+1=\omega+(1+\omega)+1=\omega+\omega+1.$$ Similarly, we have to keep straight the difference between $\omega+\omega,$ which is $\omega2$, and $2\omega$. So the answer, written most snappily, is $\omega2+1$.
