= Solution
Take $u\in L\setminus K$, whose <irreducible polynomial> is $t^2+ct+d$. Since the characteristic is not two, $x=2u+c$ satisfies $x^2=c^2-4d=a\in K$ and $L=K(x)$. Irreducibility makes $a$ nonzero and nonsquare. If another <quadratic extension> is $K(y)$ with $y^2=b$, an isomorphism sends $x$ to $s+ty$, with $t\ne0$. Squaring gives $a=s^2+t^2b+2sty$. Linear independence of $1,y$ forces $s=0$, so $a=t^2b$ and $b/a$ is a square. Conversely, if $b/a=q^2$, sending $x$ to $y/q$ gives an isomorphism. If $b$ itself is a square, the target is just $K$ and cannot be isomorphic; the square-ratio condition also fails.
The <multiquadratic field> $F$ is the <splitting field> of $\prod_i(t^2-a_i)$, a product of separable quadratic polynomials: each has distinct roots because $a_i\ne0$ and the characteristic is not two. Even if factors repeat, this splitting field is a separable normal extension, and therefore Galois. Each automorphism sends $x_i$ to $\pm x_i$, and these signs determine it, embedding its <Galois group> into $(\mathbb Z/2)^n$. Every <subgroup> of this <elementary abelian group> is $(\mathbb Z/2)^m$ for some $m\le n$.
For a quadratic intermediate field, its fixing <subgroup> $H$ has index two and is the kernel of a nontrivial <one-dimensional character> $\chi:G\to\{\pm1\}$. The coordinate sign characters span the dual of $G$: a <linear functional> on the subspace $G\subseteq\mathbb F_2^n$ extends to the ambient space by extending a basis. Hence $\chi=\prod_{i\in I}\chi_i$ for some subset $I$. Set $y=\prod_{i\in I}x_i$. Then $\sigma(y)=\chi(\sigma)y$, so $y$ is fixed by $H$ but not by all of $G$, and $y^2=\prod_{i\in I}a_i\in K$. Thus $K(y)$ is quadratic and equals the <fixed field> of $H$. This constructs all <quadratic subfields of a multiquadratic extension>:
$$
\boxed{L=K\left(\prod_{i\in I}x_i\right)}.
$$
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