Perhaps lets start with the relations between surjections, entire relations, and indexed families of nonempty sets.
It turns out these three notions are equivalent.
Suppose $f:A\to B$ is a surjection. Define a relation $R:B\to A$ by $bRa\iff f(a)=b$. Since $f$ is surjective, this relation is entire. On the other hand, if $b\in B$, define $A_b = f^{-1}(\{b\})$, since $f$ is a surjection, each $A_b$ is nonempty, so we have a family of (disjoint) nonempty sets indexed by $B$.
Now suppose we have an entire relation $R:B\to A$. Define $A_b = \{a\in A: bRa\}$, which gives a family of nonempty sets indexed by $B$, since $R$ is entire. Finally, define $$A'=\bigsqcup_{b\in B} A_b,$$
and $f:A'\to B$ by $f(a,b)=b$.
Lastly, suppose we start with a family of nonempty sets indexed by $B$, $A_b$. Then again, we define $A'=\bigsqcup_{b\in B} A_b$, and $f:A'\to B$ by $f(a,b)=b$, which is surjective, since all the $A_b$ are nonempty. On the other hand, we can define an entire relation $R:B\to A'$ by $b R (a,b)$. (Or we could take $A=\bigcup_{b\in B} A_b$ and $R:B\to A$ by $bRa \iff a\in A_b$.)
Choice
One version of the axiom of choice says that if $A_b$ is a family of nonempty sets indexed by $B$, then there is a function
$$g:B\to A'= \bigsqcup_{b\in B} A_b$$
such that $fg=1_B$, where $f:A'\to B$ is the surjective function constructed above.
$g$ is called a choice function.
Now the relationship between the statements of choice in the question is the following:
The following are equivalent
1. Choice (as stated just now)
2. Every surjective function has a right inverse.
3. Every set is projective
4. Every entire relation contains a function
Proof
(1) $\implies$ (2): Given a surjective function $f:A\to B$, and applying choice to the family of sets $A_b=f^{-1}(b)$, we get a function $g:B\to A$ such that $fg =1_B$.
(2) $\implies$ (3): Suppose $f:A\to B$ is surjective, and $h:X\to B$ is any map of sets.
To show that all sets are projective, it suffices to show that we can always lift $h$ to a map $\tilde{h}:X\to A$. However, if $g:B\to A$ is a left inverse, then we can take $\tilde{h}= gh$, since then $f\tilde{h}=fgh=h$.
(3) $\implies$ (1): Suppose $A_b$ is a family of nonempty sets. Then $f : A'\to B$ is surjective, and $B$ is projective, so we can lift $1_B$ along $f$ to a map $g:B\to A'$ such that $fg=1_B$, which is the statement of choice.
(4) $\implies$ (2): If $f:A\to B$ is surjective, and $R:B\to A$ is the entire relation constructed above, and $g:B\to A$ is a function contained in $R$, then by definition,
$bRg(b)$, which means that $fg(b)=b$, so $g$ is a right inverse to $f$.
(1) $\implies$ (4): If $R : B\to A$ is an entire relation, then we defined a family of nonempty subsets $A_b=\{a\in A: bRa\}$. Letting $\tilde{g}:B\to A'$ be a choice function for this family, we have $\tilde{g}(b) = (a,b)$ for some $a$ with $bRa$, and we define $g:B\to A$ by $g(b)=a$, which gives a function contained in $R$.
$\blacksquare$
The relations of the statements in your question
(1) is the definition of projective, which is used in statement (5).
I just showed (2) and (4) are equivalent to choice.
(6) is equivalent to saying that any family of nonempty sets indexed by $B$ has a choice function, so it's choice for sets indexed by that set.
Best Answer
The definitions are equivalent. Here is the first argument I could come up with. There may well be a simpler one using the full power of the topos assumption; the below works in any well-pointed category with a strict initial object in which every epimorphism is effective.
In a well-pointed topos, if an object $X$ has no points, then we know that $X$ has at most one morphism to any object $Y$. In other words, the unique morphism $q:0\to X$ from the initial object is an epimorphism. The dual property, of $X$ being subterminal, is widely used, but I'm not sure this one has a standard name. "Quot-initial"?
Anyway, in any topos, every epimorphism is effective. The linked nLab article cites Mac Lane-Moerdijk, IV.7.8, for this result. Thus $q$ is the quotient map for some equivalence relation $R\rightarrowtail 0\times 0$ on the initial object $0$. But since toposes have strict initial objects, we find both $0\times 0\cong 0$ and (thus) $R\cong 0$. That is, there is a unique equivalence relation on $0$ and thus $q$ is isomorphic to the identity map in the slice category under $0$.