By definition, Galois group $\operatorname{Gal}(L/K)$ consists of all $K$-linear automorphisms of $L$, where $K$-linearity is in the sense of linear algebra. For your example, you have $\mathbb Q$-linear maps acting on $L = \mathbb Q(\sqrt 2, \sqrt 3, \sqrt 5)$ and $L$ is $\mathbb Q$-algebra generated by set $\{1,\sqrt 2, \sqrt 3, \sqrt 5\}$, and that means that we only need to know where $\{1,\sqrt 2, \sqrt 3, \sqrt 5\}$ are sent to fully determine automorphism of $L$. $\mathbb Q$-linearity precisely means that $\varphi(1) = 1$ for any such automorphism $\varphi$. But where do other roots go? Let's say we want to check where $\alpha$ goes, and let $f$ be it's minimal polynomial over $\mathbb Q$. Then $0 = \varphi(f(\alpha)) = f(\varphi(\alpha))$ thus $\varphi(\alpha)$ is the root of the same polynomial as $\alpha$ ($f$ and $\varphi$ commute because $f$ is polynomial and $\varphi$ is ring homomorphism). So, in your case, $\sqrt 2$ goes to $\pm \sqrt 2$ because minimal polynomial of $\sqrt 2$ is $x^2 - 2$ and similarly, $\sqrt 3$ maps to $\pm \sqrt 3$ and $\sqrt 5$ maps to $\pm\sqrt 5$. Thus, there are only $8$ possible automorphisms, so Galois group is group with $8$ elements.
Now, for the subgroups, Fundamental theorem of Galois theory tells us that any subgroup of Galois group corresponds to a subextension of fields. How do you get this field? Well, for a subgroup $H$, intermediate field $L'$ is given by set $\{ x\in L \ | \ \varphi(x) = x,\ \forall \varphi\in H\}$. And yes, to your question, subgroup spanned by $\varphi = [\sqrt 2\mapsto \sqrt 2, \sqrt 3\mapsto -\sqrt 3, \sqrt 5\mapsto -\sqrt 5]$ is $\{\mathrm{id}, \varphi\}$ because $\varphi^2 = \mathrm{id}$.
Timbuc already wrote explicit basis for $L$ and $\varphi$ fixes $\{1,\sqrt 2, \sqrt {15}, \sqrt {30}\}$ which corresponds to field $\mathbb Q(1,\sqrt 2, \sqrt {15}, \sqrt {30}) = \mathbb Q(\sqrt 2, \sqrt 15)$ since other generators can be obtained by multiplication.
Best Answer
$G$ is the Galois group of the extension $E:F$ where $E$ is the splitting field of $f$ over $F$. Now, that means that $E=F(a_1,\cdots, a_n)$, where the $a_i$ are the roots of $F$. Now, you can directly prove that every element of the Galois group $G$ is determined by what it does on the roots, and that every root must be mapped to a root. So, there is an injective map $G\to S_{\{a_1,\cdots ,a_n\}}$. This functions is easily seen to be a homomorphism of groups, thus an embedding of the Galois group $G$. Now, identify the codomain group with $S_n$, and you obtain that the Galois group can be identified with a subgroup of $S_n$.