I am trying to find $[\mathbb{Q}(\sqrt{p},\sqrt{q}),\mathbb{Q}]$ where $p,q$ are distinct primes in $\mathbb{N}$.
I know that the elements of $\mathbb{Q}(\sqrt{p})$ can be obtained by considering $F[\sqrt{p}]=\{f(\sqrt{p}): f \in \mathbb{Q}[x]\}$ and as the minimal polynomial of $\sqrt{p}$ is $x^2-p$ from the division theorem, we obtain that the form of the elements of $\mathbb{Q}(\sqrt{p})$ is $a+b\sqrt{p}$.
Now I need to do it again, so the elements of $\mathbb{Q}(\sqrt{p})(\sqrt{q})$ can be obtained by considering $G[\sqrt{q}]=\{f(\sqrt{q}): f \in \mathbb{Q}(\sqrt{p})[x]\}$ where $\mathbb{Q}(\sqrt{p})[x]$ are the polynomials with coefficients in $\mathbb{Q}(\sqrt{p})$, I think this should yield polynomials such as:
$$(a_{n} +b_{n}\sqrt{p})X^{n}+(a_{n-1} +b_{n-1}\sqrt{p})X^{n-1} + \dots$$
Now I am a bit confused: How do I find the minimal polynomial of $\sqrt{q}$ with coefficients in $\mathbb{Q}(\sqrt{p})$?
I tried to use this proposition:
From this, we know that $[\mathbb{Q}(\sqrt{p},\sqrt{q}),\mathbb{Q(\sqrt{p})}]$ has degree $1$ or $2$ but supposing that $[\mathbb{Q}(\sqrt{p},\sqrt{q}),\mathbb{Q(\sqrt{p})}]=1$ and obtaining a contradiction seemed harder. If that is true: It means that $\mathbb{Q}(\sqrt{p}, \sqrt{q})=\mathbb{Q(\sqrt{p})}$ which implies that $\sqrt{q} \in \mathbb{Q(\sqrt{p})}$ and hence:
$$\sqrt{q}=a+b{\sqrt{p}}$$
But I was not able to find a contradiction on the existence of the coefficients $a,b$.
Your idea is good! Indeed, $\sqrt q\notin\mathbb Q(\sqrt p)$ and we can in fact derive a contradiction assuming that
$$\sqrt q=a+b\sqrt p$$
for some $a,b\in\mathbb Q$. First note that $a\ne0$ as writing $b=\frac xy$ and squaring both sides would give
$$q=b^2p\implies qy^2=x^2p$$
and hence a contradiction to unique factorization as the occuring factors $p,q$ on the LHS and the RHS differ. Now observe that
\begin{align*} \sqrt q=a+b\sqrt p\implies(\sqrt q-b\sqrt p)^2=a^2\iff \sqrt{pq}=\frac{a^2-b^2p}{2b} \end{align*}
given that $b\ne0$ (which is clear as $\sqrt p$ is irrational). Hence, if $\sqrt q\notin\mathbb Q(\sqrt p)$ then $\sqrt{pq}$ is rational. But a variation of the classical argument for square roots of prime numbers shows that $\sqrt{pq}$ is irrational as well and hence the desired contradiction is found.
This generalizes nicely to finite sets of pairwise distinct primes, i.e. for
$$p_1,\dots,p_n\quad \text{we have}\quad [\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_n}):\mathbb Q]=2^n$$
and the minimal polynomials at each step are just $X^2-p_i$. The proof follows the same argument as above using induction on the number of primes.
A more general note: if you have a field extenstion of the form $K(\alpha,\beta)$ and minimal polynomials $f=\min_K(\alpha)$, $g=\min_K(\beta)$ you can always first check whether or not, say, $g$ is irreducible in $K(\alpha)$ viewed as element of $K(\alpha)[X]$. If so, it will again be the minimal polynomial of $\beta$ in $K(\beta)$ and $[K(\alpha,\beta):K(\alpha)]=[K(\beta):K]$. This strategy might bypass a tedious calculation of minimal polynomials.