Computing class group of $\mathbb Q(\sqrt{6})$

There are indeed no elements of norm $2$, but there are elements of norm $-2$.

Where you went wrong is where you overlooked the Diophantine equation $x^2 - 6y^2 = -2$. The fundamental unit of this ring is $5 + 2 \sqrt 6$, which has a norm of 1, not $-1$. This means that if $x^2 - 6y^2 = n$ has no solutions, it does not rule out that $x^2 - 6y^2 = -n$ might.

And indeed $x^2 - 6y^2 = -2$ does have infinitely many solutions, with the most obvious one being $x = 2$, $y = 1$, which you yourself later discovered. Then $(2 + \sqrt 6)(5 + 2 \sqrt 6)$ gives another, and from there you can figure out how to get as many as you want.

More importantly to your query, however, $(2 - \sqrt 6)(2 + \sqrt 6) = -2$, and since $-2 \in \langle 2 \rangle$, combined with the fact that $(2 + \sqrt 6)(3 - \sqrt 6) = \sqrt 6$, we find that both $\langle 2 \rangle$ and $\langle \sqrt 6 \rangle$ are properly contained in $\langle 2 + \sqrt 6 \rangle$, which is not only a principal ideal but a prime ideal after all.

You were right that 2 ramifies, as we have $\langle 2 \rangle = \langle 2 + \sqrt 6 \rangle^2$. In general, if $\mathbb Q(\sqrt{pq})$ (with $p$ and $q$ distinct primes) is a principal ideal domain, you can find factorizations for $\pm p$, $\pm q$ in that domain, and consequently that $\sqrt{pq}$ is reducible.