Definition of "CM-field"

Yes.

• The definition: $K$ doesn't have any real embedding and there is some subfield such that $[K:F]=2$ and every complex embedding sends $F$ to $\Bbb{R}$.

• $[K:F]=2$ gives that $K=F(\sqrt{a})$ for some $a\in F$. For each complex embedding $\sigma\in Hom_\Bbb{Q}(K,\Bbb{C})$ then $\sigma(K)=\sigma(F)(\sigma(\sqrt{a}))$.

$F$ is totally reals means that $\sigma(F)\subset \Bbb{R}$.

So the complex conjugaison acts on $\sigma(K)$ by sending $\sigma(\sqrt{a})$ to $\overline{\sigma(\sqrt{a})}$. Since also $a\in F$ it means that $ \overline{\sigma(\sqrt{a})}=-\sigma(\sqrt{a})$. Whence $\sigma(F) = \sigma(K)\cap \Bbb{R}$ and $$F=\sigma^{-1}(\sigma(K)\cap \Bbb{R})$$

• As you see $F$ doesn't depend on the chosen $\sigma$, its uniqueness is implied by the definition of CM-field, as well as that any complex conjugaison gives an automorphism of $K$, and that there is in fact only one complex conjugaison on $K$.

If I understand your question correctly, then no. A number field $K$ can be a CM-field without specifying the base field over which it is a quadratic imaginary extension. In fact, we say $K$ is a CM-field if there simply exists some totally real subfield $F$ for which $K$ is an imaginary quadratic extension.