I am working on the question of why the ideal $(2,\sqrt{-6})$ is not a principal ideal in $\mathbb{Q}(\sqrt{-6})$, but becomes one in $\mathbb{Q}(\sqrt{-6},\sqrt{2})$.
To prove that it is not principal in $\mathbb{Q}(\sqrt{-6})$, I first computed the norm of this ideal:
$$N((2,\sqrt{-6}))=|\mathbb{Z}[\sqrt{-6}]/(2,\sqrt{-6})|=|\mathbb{Z}[T]/(2,T,T^2+6)|=|\mathbb{F}_2|=2.$$
Then I supposed that there exists $\alpha\in\mathbb{Q}(\sqrt{-6})$, $\alpha=a+b\sqrt{-6}$ with $a,b\in\mathbb{Q}$, so that $(2,\sqrt{-6})=(\alpha)$ in $\mathbb{Q}(\sqrt{-6})$. Without loss of generality I assumed that $a$ and $b$ already are in $\mathbb{Z}$ (if not, multiply by a suitable unit). Then the norm of the ideal is
$$N((\alpha))=N_{\mathbb{Q}(\sqrt{-6})/\mathbb{Q}}(\alpha)=(a+b\sqrt{-6})(a-b\sqrt{-6})=a^2+6b^2.$$ But the equation $a^2+6b^2=2$ can't be solved in $\mathbb{Z}$. Therefore $(2,\sqrt{-6})$ is not a principal ideal in $\mathbb{Q}(\sqrt{-6})$.
Now the next step is to find $\beta\in\mathbb{Q}(\sqrt{-6},\sqrt{2})$, so that $(2,\sqrt{-6})=(\beta)$ in $\mathbb{Q}(\sqrt{-6},\sqrt{2})$. How can I find this $\beta$?
You've done the hard part!
First note that $\mathbb Q(\sqrt{-6},\sqrt2)=\mathbb Q(\sqrt{-3},\sqrt2)$. Observe that $\sqrt{2}$ divides both $2$ and $\sqrt{-6}$ in $\mathbb Q(\sqrt{-6},\sqrt2)$, since $$2=\sqrt2\cdot\sqrt 2\\\sqrt{-6}=\sqrt2\sqrt{-3}.$$
Hence $(\sqrt{-6},2)\subset(\sqrt2)$.
Moreover, $$\sqrt{-6}\cdot\sqrt{-3}=\sqrt{18}=3\sqrt2\text{ and}\\\sqrt 2 = 3\sqrt2-2\sqrt2,$$ so $$\sqrt2 = (\sqrt{-6}\cdot\sqrt{-3})-(2\cdot\sqrt2)\in(\sqrt{-6},2).$$ Hence $(\sqrt{-6},2)\supset(\sqrt2)$.