How could I express this result better, quantificationally? For a pre-specified $a$,
$\exists B \subseteq \mathbb{R}^{n+k} \ \ \exists f:B \to \mathbb{R}^k \ \ \forall x \in B: a \in B \ \text{and} \ F(x, f(x)) = 0$
I just dislike how we are 'picking' an $x$ from $B$ before saying that $a \in B$.
The following equivalences $$\forall x \in X(\phi \land \psi) \equiv \phi \land (\forall x \in X\,\psi)$$
$$\exists x \in X(\phi \land \psi) \equiv \phi \land (\exists x \in X\,\psi)$$
are valid providing $x$ does not occur free in $\phi$. In your example, you can use these to move the subformula $a \in B$ to just under the quantification over $B$:
$$\exists B \subseteq \mathbb{R}^{n+k} (a \in B \land (\exists f \in B \to \mathbb{R}^k \ \ \forall x \in B \ F(x, f(x)) = 0)).$$
Note that this is equivalent to what you had, but I agree with you that it reads better.
[Aside: I've converted your mixture of uses of $\in$ and $:$ into a more common convention. Feel free to convert back to the conventions of your course or book.]