Let $f(x)=x^4-6x^2+4 \in \mathbb{Q}[x]$.
The Galois group is $\mathrm{Gal}(L/K) \simeq \mathbb{(Z/2Z)^2}$, but I don't know how to find it.
I know that $\mathbb{Q}(\sqrt{3+\sqrt{5}})$ is a splitting field of $f(x)$ over $\mathbb{Q}$.
$\mathrm{Gal}(L/K)$ acts transitively on the roots of $f(x)$, so there exist $\sigma_1, \sigma_2, \sigma_3$ and $\sigma_4$ with $\sigma_1(\sqrt{3+\sqrt{5}})=\sqrt{3+\sqrt{5}}=\mathrm{id}, \sigma_2(\sqrt{3+\sqrt{5}})=-\sqrt{3+\sqrt{5}}, \sigma_3(\sqrt{3+\sqrt{5}})=\sqrt{3-\sqrt{5}}$ and $\sigma_4(\sqrt{3+\sqrt{5}})=-\sqrt{3-\sqrt{5}}$
So $\sigma_i^2=\mathrm{id}$ and $\mathrm{Gal}(L/K) \simeq \mathbb{(Z/2Z)^2}$
This is what I don't understand. I see that $\sigma_2^2=\sigma_2 \circ \sigma_2=\sigma_1$, but $\sigma_3^2=\sigma_3 \circ \sigma_3 =0,7639 \neq \mathrm{id}$ aswell as $\sigma_4^2=\sigma_4 \circ \sigma_4 = 0,7639 \neq \mathrm{id}$.
How to compute it to get $\sigma_3^2=\sigma_4^2=\mathrm{id}$?
A quick way to compute the Galois group of a quartic like this is to use the discriminant
$$\Delta(P) = 2560X^3 - bX^2 + (ac−4d)X - (a^2d+c^2−4bd)0$$ which is a square, and the cubic resolvent
$$R_3(X) = y^3 - by^2 + (ac−4d)y - (a^2d+c^2−4bd) \\= y^3 + 6 y^2 - 16 y - 100$$
which has no rational roots in this case.
These two facts together tell us the Galois group is $C_2 \times C_2$.
See https://kconrad.math.uconn.edu/blurbs/galoistheory/cubicquartic.pdf for details.
Alternatively (inspired by the insight in the comment by Jyrki Lahtonen), write the polynomial as $$X^4 - 2 \cdot 3 X^2 + 4$$ and put $X = \sqrt{2} Y$ so that we have $$4 Y^4 - 4 \cdot 3 Y^2 + 4 = Y^4 - 3 Y^2 + 1 = (Y^2 - Y - 1)(Y^2 + Y - 1)$$
Both quadratics have discriminant 5. A root of the first polynomial may be written as $Y = \frac{1 + \sqrt{5}}{2}$ and so we have an expression $X = \sqrt{2}\frac{1 + \sqrt{5}}{2}$.
All four roots are got by similar expressions (conjugate the square roots) and so the splitting field is $$\mathbb Q(\sqrt{5}, \sqrt{2})$$ and therefore the Galois group is $V_4$.