This article proves that biquadratic extensions correspond precisely to Galois extensions with Galois group isomorphic to the Klein $4$ group $V_4$ (at least if the characteristic of the base field is not $2$ . More precisely,

Proof. Suppose first that condition (1) holds. Then $[F(\sqrt{D_1}):F]=[F(\sqrt{D_2}):F]=2$ since neither $D_1$ nor $D_2$ is a square in $F$ Now obviously $$[K:F]=[F(\sqrt{D_1},\sqrt{D_2}):F(\sqrt{D_1})][F(\sqrt{D_1}):F]\leq 4$$ and so $[K:F(\sqrt{D_1})]\leq 2$ If $K=F(\sqrt{D_1})$ then $\sqrt{D_2}\in F(\sqrt{D_1})$ so $\sqrt{D_2}=a+b\sqrt{D_1}$ and $D_2=a^2+b^2D_1+2ab\sqrt{D_1}$ Thus $a=0$ or $b=0$ If $b=0$ then $D_2$ is a square. If $a=0$ then $D_1D_2=b^2D_1^2$ is a square. In any case, this is a contradiction. Thus $K$ is a quadratic extension of $F(\sqrt{D_1})$ So $[K:F]=4$ But $K$ is the splitting field for $(x^2-D_1)(x^2-D_2)$ since the splitting field must contain both square roots, and the polynomial obviously splits in $K$ so $G=\Gal(K/F)$ has four elements

Now assume that condition (2) holds. Since $\Gal(K/F)\cong V_4$ there must be three intermediate subfields $E_1, E_2, E_3$ between $F$ and $K$ of degree $2$ over $F$ corresponding to the three subgroups of $V_4$ of order $2$ Thus each of these is a quadratic extension. Suppose $E_1=F(\sqrt{D_1}), E_2=F(\sqrt{D_2})$ where neither $D_1$ nor $D_2$ is a square in $F$ The fact that $E_1\neq E_2$ implies as above that $D_1D_2$ is also not a square in $F$ (in fact $E_3=F(\sqrt{D_1D_2})$ Thus $E_1E_2\supsetneq E_1, E_2$ and is of degree $4$ over $F$ so $K=E_1E_2 =F(\sqrt{D_1},\sqrt{D_2})$