I understand why $(\omega+1)\cdot2 = \omega\cdot2+1$ and why $(\omega+1)\cdot\omega = \omega^2$
What I am struggling with is something along the lines of:
$(\omega\cdot3+4)\cdot3$ which I think is = $\omega\cdot9+4$
And:
$(\omega\cdot3+4)(\omega\cdot3)$ which I think is = $\omega^2\cdot9$
Am I correct?
You have evaluated $(\omega\cdot 3+4)\cdot 3$ correctly. However, $(\omega\cdot 3+4)\cdot(\omega\cdot 3)$ is not correctly evaluated. I do not know which step of your evaluation is incorrect, but I guess you think $(\omega\cdot 3)^2$ is $\omega^2\cdot 9$.
This is not true, however, as $$(\omega\cdot 3)\cdot (\omega\cdot 3) = \omega\cdot (3\cdot\omega)\cdot 3 = \omega^2\cdot 3.$$ The core part of the evaluation is $3\cdot\omega=\omega$ that can be proven in the way of proving $3+\omega =\omega$ (and the linked question in your comment.)
We would think evaluating $(\omega\cdot 3+4)\cdot(\omega\cdot 3)$ is not too different from that of $(\omega\cdot 3)\cdot (\omega\cdot 3)$. The problem is, we have extra $+4$ and we cannot apply the distribution law in the case. However, the addition in the middle of the expression would disappear when adding $(\omega\cdot 3+4)$ many times, so we can guess the answer is $\omega^2\cdot 3$.
You can make our discussion concretely by following the steps of the proof you have linked; that is, use the transfinite induction for $\beta\le \omega\cdot 3$ to prove $$(\omega\cdot 3+4)\cdot\beta = \begin{cases} \omega\cdot (3\cdot\beta) &\text{if $\beta$ is a limit ordinal,} \\ \omega\cdot (3\cdot\beta) + 4 & \text{if $\beta$ is a successor ordinal.}\end{cases}$$