Why do we have a basis? A corollary that is in my book that I think is relevant to my question is:

If E is an extension field of F, $\alpha \in E$ is algebraic over F,
and $\beta \in F(\alpha)$, then $\deg(\beta,F)$ divides
$\deg(\alpha,F)$.

I have two questions regarding the green and red line in the example under, the red line is my main concern:
green question: Does what they say here follow directly from the corollary above and the fact that 4 does not divide 2?
red question: Why does it now follow that $\{1,\sqrt{3}\}$ is a basis for our vector-space?
I must admit that even just $(\mathbb{Q}(\sqrt{2}))(\sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$, is a little hard to grasp. But as I see it consists of elements $a_0+a_1\sqrt{3}$, or $a_0+a_1\sqrt{3}+a_23$, where $a_0,a_1,a_2 \in \mathbb{Q}(\sqrt{2})$?(is this correct?) The reason I have two options here is that I do not know the dimension of $(\mathbb{Q}(\sqrt{2}))(\sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$.

 A: If $\alpha/F$ has deg $n$ then $\alpha$ cannot exist in an extension of $F$ of deg $<n$. One could argue this using vector space dimensions (if $F(\alpha)\subseteq L$ then $n\le \dim_FL$), although you can also employ the transitivity of field degrees if you want (seems only slightly stronger than necessary). Either way, we may conclude that since $\sqrt{2}+\sqrt{3}$ has deg $4$ over $\Bbb Q$ it cannot exist in $\Bbb Q(\sqrt{2})$ as that's only deg $2$ over $\Bbb Q$.
You are correct that you have two options, but you are wrong about what those options are. Note that every number of the form $a_0+a_1\sqrt{3}+a_2\sqrt{3}^2$ (with $a_0,a_1,a_2\in\Bbb Q(\sqrt{2})$) directly simplifies to the form $b_0+b_1\sqrt{3}$ (with $b_0,b_1\in\Bbb Q(\sqrt{2})$), since $b_0:=a_3+3a_2\in\Bbb Q(\sqrt{2})$. So there is no point to having a superfluous $a_23$. You should be aware of the very important Fact below to understand this more.
The actual two options are that elements of $\Bbb Q(\sqrt{2})(\sqrt{3})$ look like $a_0+a_1\sqrt{3}$ ($a_0,a_1\in\Bbb Q(\sqrt{2})$) or else they simply look like $a\in\Bbb Q(\sqrt{2})$ if it were to so happen that $\Bbb Q(\sqrt{2},\sqrt{3})=\Bbb Q(\sqrt{2})$ (which is equivalent to $\sqrt{3}\in\Bbb Q(\sqrt{2})$). It is perfectly natural to not a priori know the dimension of $\Bbb Q(\sqrt{2},\sqrt{3})$ over $\Bbb Q(\sqrt{2})$ - part of the exercise is proving it has dimension $2$.
Fact: If $\alpha/F$ has deg $n$ then $F(\alpha)/F$ has a vector space basis $\{1,\alpha,\cdots,\alpha^{n-1}\}$.
Proof. The set $\{1,\alpha,\cdots,\alpha^{n-1}\}$ is linearly independent over $F$, since any linear relationship among them would be a polynomial relation of degree $<n$ satisfied by $\alpha$ and that would contradict the fact $\alpha$ has degree $n$ over $F$. Since $|\{1,\alpha,\cdots,\alpha^{n-1}\}|=n=\dim_FF(\alpha)$ and this set is linearly independent over $F$, we conclude that it is a basis.
Since $\Bbb Q(\sqrt{2},\sqrt{3})$ and $\Bbb Q(\sqrt{2}+\sqrt{3})$ both have deg $4$ over $\Bbb Q$, and one contains the other, they must be equal, so $[\Bbb Q(\sqrt{2},\sqrt{3}):\Bbb Q]=4$. But since $[\Bbb Q(\sqrt{2}):\Bbb Q]=2$, by transitivity of degrees we are forced to conclude $[\Bbb Q(\sqrt{2})(\sqrt{3}):\Bbb Q(\sqrt{2})]=2$ so $\{1,\sqrt{3}\}$ is a basis for $\Bbb Q(\sqrt{2},\sqrt{3})/\Bbb Q(\sqrt{2})$ (see Fact).
