I noticed $(\omega+2)*2$ had some pretty weird results: \begin{align} (\omega+2)*2 & = (\omega+2)+(\omega+2) \\ & = \omega+(2+\omega)+2 \\ & = \omega+\omega+2 \\ & = \omega2+2 \\ \end{align}
So I wonder how $(\omega+2)²$ behaved:
$$(\omega+2)^2 = (\omega+2)*(\omega+2)$$
but I don't know how the order of the distributive-ity works here like:
$$(\omega+2)*(\omega+2) = (\omega+2)*\omega + (\omega+2)*2$$
or:
$$(\omega+2)*(\omega+2) = \omega*(\omega+2) + 2*(\omega+2)$$
I think $(\omega+2)*\omega = \omega^2$ because the $+2$ gets consumed by the infinite additions.
But $(\omega*(\omega+2) = \omega^2+\omega2$ because you reach $\omega^2$ and than still have some left over.
A similar explanation can be applied too: $$(ω+2)∗2=(ω+2)+(ω+2)=ω2+2$$ $$2∗(ω+2)= 2+2+...+2 \text{ ($\omega+2$ times)} = \omega+2$$
Concrete
So which of these equations is correct?
And how does distribution work on ordinals?
Ordinal arithmetic follows the distributive law on the left, that is,
$$\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma$$
In your case this means
$$(\omega+2)^2=(\omega+2)(\omega+2)=(\omega+2)\omega+(\omega+2)2=\omega^2+\omega2+2$$