How is $Gal(\mathbb Q(\sqrt{2+\sqrt{2}})/\mathbb Q)\cong \mathbb Z/4\mathbb Z$?

If the Galois group were $(\mathbb Z/2\mathbb Z)^2$, then $L$ would contain two quadratic subfields. One of them is $\mathbb Q(\sqrt 2)$; what is the other one?

In fact I claim that $L$ contains no other quadratic subfield. A quick way to see this using number theory is that if $L$ contained $\mathbb Q(\sqrt D)$ where $D$ is odd, then $L$ would be ramified above some odd prime. But the discriminant of $L$ divides the discriminant of $x^4-3x^2+2$ which is $32$, hence $disc(L)$ is a power of $2$. So $L$ cannot be ramified above any odd prime. You can definitely come up with an elementary proof, though.

Your mistake is in incorrectly presenting the Galois group. How do you know that that's what the Galois group does to the roots?

$\def\a{{\alpha}}\def\t{{\tau}}\def\Gal{\operatorname{Gal}}\def\Q{{\mathbb Q}}$Your mistake was the assumption that $\t(\sqrt{2})=\sqrt{2}$; $\t(\sqrt{2})=-\sqrt{2}$:

$$ \a_1 = \sqrt{2+\sqrt{2}}, \a_3 = \sqrt{2-\sqrt{2}}\\ \begin{align} \t(\a_3)&=\t\left(\frac{\sqrt{2}}{\a_1}\right)\\ &=\t\left(\frac{\color{red}{\a_1^2-2}}{\a_1}\right)\\ &=\frac{\a_3^2-2}{\a_3}\\ &=\frac{-\sqrt{2}}{\a_3}\\ &=-\a_1=\a_2 \end{align} $$

And since the $\a_1$ then has order $> 2$, we know that $\t(\a_2)=\a_4, \t(\a_4)=\a_1$. Note that we don't need to worry if $\t(2)=2$ since $\t\in G$ so $\t$ fixes $\Q$