The way we define $\mathbb{Q}(\sqrt{2})$ is via the following set,
$F(u) := \{ f(u)/g(u):f(u),g(u)\in F[x],g(u)\neq 0 \}$
So, I am then confused on how one defines the set $\mathbb{Q}(\sqrt{2})(\sqrt{3})$. And how is this set functionally different from $\mathbb{Q}(\sqrt{2},\sqrt{3})$? How would one even define $\mathbb{Q}(\sqrt{2},\sqrt{3})$? (It was not defined explicitly with commas anywhere in my lecture).
The notation I'm familiar with is that, given a field $F$ and a specific element $\alpha$ of some field extension of $F$ (say, $K$), then $F(\alpha)$ denotes the smallest field containing both $\alpha$ and $F$. That is,
$$F(\alpha) := \bigcap_{\substack{\text{fields $L$} \\ \alpha \in L \\ F \subseteq L}} L$$
One may extend this in the obvious way to $F(\alpha_1,\cdots,\alpha_n)$, the smallest field containing all of $F$ as well as all of the $\alpha_i$.
One may show that $F(\alpha,\beta) = (F(\alpha))(\beta)$.
Given an indeterminant $x$ and field $F$, however, $F(x)$ denotes all rational functions over $F$, i.e.
$$F(x) := \left\{ \frac{a(x)}{b(x)} \, \middle| \, a,b \in F[x], b \ne 0 \right\}$$
and likewise
$$F(x_1,\cdots,x_n) := \left\{ \frac{a(x_1,\cdots,x_n)}{b(x_1,\cdots,x_n)} \, \middle| \, a,b \in F[x_1,\cdots,x_n], b \ne 0 \right\}$$
Given $\alpha$, there is a natural equivalence between $F(\alpha)$ as defined above, and $F(x)$ when each rational function is evaluated at $x=\alpha$.
Frankly, this is something I've always found a bit confusing, but this answer somewhat cleared things up for me.
It may be simplest to consider: