For clarity, consider first the general problem of embedding a finite galois extension $K/k$, with group $G$, into a finite overextension $E/F/k$, and give a necessary and sufficient for $E/k$ to be galois. Almost by the definition of normality, such a NSC is that all the extensions of the $k$-automorphisms $s$ of $F$ (i.e. of the elements of $G$) to $k$-embeddings $\hat s$ of $E$ in a separable closure of $k$, should stabilize $E$. If $E$ is given by $E=F(\alpha)$, this means that all the conjugates $\hat s(\alpha)$ belong to $E$.
In your particular case, $k=\mathbf Q$, $F=\mathbf Q (\sqrt 2, \sqrt 3)$ is a biquadratic field, and $E=F(u)$ with $u^2=v=(2-\sqrt 2)(9-5\sqrt 3) \in F$. The group $G$ is obviously $\cong C_2\times C_2$, generated by $s$ s.t. $s(\sqrt 2)=-\sqrt 2, s(\sqrt 3)=\sqrt 3$, and $t$ defined likewise by permuting $2$ and 3. Because the quadratic extension $F(\sqrt v)$ depends only on the class of $v$ mod ${F^*}^2$, the stabilization condition above readily amounts to $s(v)/v$ and $t(v)/v \in {F^*}^2$. This is easily checked, so that $E/\mathbf Q$ is galois of degree $8$.
There are only $3$ possibilities for a group of order $8$ admitting a quotient $\cong C_2\times C_2$, which are $C_2\times C_2\times C_2$, the dihedral $D_8$, or the quaternionic $H_8$. The first case can be immediately discarded on looking at $u$ (why ?). To distinguish between the $2$ latter cases, a pedestrian way is to extend $s$ and $t$ explicitly to $E$ and compute the (non)commutation relations between $\hat s$ and $\hat t$. For a (possible) shortcut, see e.g. https://math.stackexchange.com/a/2046653/300700 .
