The axiom of choice states that given a collection of non-empty sets $\mathcal A=\{A_i\mid i\in A\}$ there is a function such that $F(A_i)\in A_i$. Such $F$ is called a choice function for the family $\cal A$. It should be remarked that there are plenty of instances where the axiom of choice is really needed for proving that such function exists for a certain family.
So now suppose that $F\colon A\to P(B)$ as described. By the covering property we have that for every $b\in B$ the set $A_b=\{a\in A\mid b\in F(a)\}$ is non-empty. Therefore the family $\{A_b\mid b\in B\}$ is a family of non-empty sets, and by the axiom of choice there is a choice function defined for it.
Let $g$ be a choice function from the family $\{A_b\mid b\in B\}$, we have that $g(A_b)\in A_b$. Let now $G(b)=g(A_b)$, and we have that $G(b)\in A_b$ so by definition $b\in F(G(b))$, as wanted.
It is impossible to solve this question with a weak choice principle (namely an assertion implied by the axiom of choice, but not equivalent to the axiom of choice).
To see this note that if $\{A_i\mid i\in I\}$ is a family of non-empty sets we can take the function $F\colon\bigcup_{i\in I}A_i\to P(I)$ defined as $F(a)=\{i\in I\mid a\in A_i\}$. Then $F$ covers $I$ because for every $i\in I$ we have some $a\in A_i$ and then $i\in F(a)$ for such $a$.
Suppose that $G\colon I\to\bigcup_{i\in I}A_i$ is such that $F(G(i))=i$ then $G(i)\in A_i$ and therefore $G$ is a choice function for $\{A_i\mid i\in I\}$.
So if the principle of "splitting a covering function" holds every family of non-empty sets has a choice function, i.e. the axiom of choice holds.