The key in these kinds of proofs is to find a particular instance of the proposition you are assuming that will yield the one you want. So in order to prove the second form from the first, you start with $I$ and $H$, and want to "cook up" some $R$ for which the $G$ will yield what you want. And conversely.
You are sort of on the right track, but the problem is that you are trying to apply each form to the information from the other, instead of cooking up an appropriate set to apply the "other" form in each case. For instance, in your second proof, you shouldn't take $H$ to be any function, you want to cook up a particular function.
Now, before we actually prove they are equivalent, let's think about them. Intuitively, why should these two "be" the Axiom of Choice? In a relation, every element in the domain is associated to many elements in the codomain. Being able to find a function "contained" in the relation is like, for each $x$ in the domain, picking one element from each of the things that are associated to $x$ to be the image of $x$. This is the Axiom of Choice (you are picking one element from each collection of "things related to $i$" for $i\in I$). As for the second, an element of $\mathop{\times}\limits_{i\in I}H(i)$ is a function from $I$ to $\cup H(i)$ such that the image of $i$ is in $H(i)$ for each $i$; again, a choice of elements from a family of sets, again Choice. Knowing how each of them relates to AC, and hence to each other, should help you figure out how to show the equivalence: after all, that equivalence should be one in which the particular "choice" that one form allows you to make is exactly the "choice" that the other form requires you to make.
For more explicit hints:
In your attempt to show the first form implies the second, you are focusing on the wrong thing; you don't want to look at $H$ as a relation, because then you will have $G=H$ ($H$ is already a function). Instead, define a relation $R\subseteq I\times \cup H(i)$ by $(i,x)$ if and only if $x\in H(i)$. (That is, $H(i)$ is a nonempty set for each $i$; associate with each $i$ all elements in $H(i)$). Now use the first form with $R$, not with $H$, because the $G$ should give you an element of the product (think of them as described above).
For the second, assume that $R$ is a relation, and you want to define a function contained in $R$ with the same domain. Let $I$ be the domain of $R$, and for each $i\in I$, let $H(i) = \{ x\in\mathrm{codom}(R)\mid iRx\}$. Use the fact that the product is nonempty to pick an element in the product (which "picks" an element from each of the sets of "things related to $i$") and use that to define the function $G$.
The point of the Axiom of Choice is that oftentimes a mathematician finds him- or herself at a point where infinitely many specific objects must be gathered up all at once in order to continue a proof/construction.
In the statement as given, we have a family of nonempty sets $\{ A_i : i \in I \}$, and we want to pick a unique representative from each $A_i$, this is our function $f : I \to \bigcup_{i \in I} A_i$ ($f$ "picks" $f(i)$ to be the unique representative from $A_i$). The Axiom of Choice says that this is unproblematic, and we can always do this.
Of course, there are certain specific instances where one can do this without appealing to the Axiom of Choice:
- if you only have to make finitely many choices; or
- if these choices can be made in a uniform manner (e.g., if I have infinitely many nonempty sets of natural numbers, I can choose the least one from each set).
More often there is no way to uniformly pick these representatives, and without appealing to some extra-logical hypothesis we cannot make the choices as required.
(There are many statements known to be equivalent to the Axiom of Choice. The most common one you see in mathematics outside of logic/set theory is Zorn's Lemma.)
Best Answer
$3\Rightarrow 2$: Suppose $X$ is a collection of nonempty sets, then $f:P(\bigcup X)\setminus\{\varnothing\}\to\bigcup X$ exists which chooses from every nonempty subset of $\bigcup X$. If we restrict $f$ to the set $X$ then we have a choice function.
$2\Rightarrow 3$: Trivial, if every non-empty collection of non-empty sets has a choice functions, given a non-empty $X$ we have that $P(X)\setminus\{\varnothing\}$ is a non-empty collection of non-empty sets, thus has a choice function.
$2\Rightarrow 1$: Trivial, if every collection has a choice function then every pairwise disjoint collection has a choice function $\{f(x)\mid x\in X\}$ is a choice set.
$1\Rightarrow 2$: Given $X$ a non-empty collection of non-empty sets, let $\{\{x\}\times x\mid x\in X\}$ be a collection of now pairwise disjoint sets. The cut set is a choice function. (You may want to show why they are pairwise disjoint, this is because if $x\neq y$ then $(\{x\}\times x)\cap(\{y\}\times y)=\varnothing$; also you might want to show why the choice set is a function, but it only meets $\{x\}\times x$ at one point... so functionality holds and it is indeed a choice function.)