This is an answer to the second question, proving Global Choice is bi-interpretable with a rank-respecting Global Wellorder. This is intuitively the same as finding a definable bijection between the two relevant hyperclasses.
Our strategy is inspired by the fact each global wellorder can be expressed as a concatenation of its restrictions to each rank of the cumulative heirarchy, the sets $V_{\alpha+1}\setminus V_\alpha$. Similarly, each global choice function can be expressed as a union of its restrictions to each rank domain. This decomposition allow us to represent our hyperclasses as products of sets, indexed by the ordinals. This works since the behavior of a choice function/wellorder within one rank is completely independent to its behavior in some other rank.
Ideally we can find, for each rank $\alpha$, a specific bijection between the corresponding factors of the above product representations. If this can be done by a single formula (parameterized by $\alpha$), then we can produce a bijection of the relevant hyperclasses using their product representation, leading to an explicit bi-interpretation. As-is, this strategy runs into numerous difficulties. Instead, we use a slightly modified strategy where the decomposition is broken up only at limit ranks.
Bijecting the factors
Our technique relies on the fact that the Schroder-Bernstein theorem is roughly constructive. That is, the following fact should be a corollary of the typical Schroder-Bernstein proof.
There's a set-theoretic formula $\phi(f,g,h)$ such that $\sf{ZF}$ proves: if $f,g$ are injections where both $\operatorname{ran}(g)\subseteq \operatorname{dom}(f)$ and $\operatorname{ran}(f)\subseteq \operatorname{dom}(g)$, then there's a unique $h$ such that $\phi(f,g,h)$, and this unique $h$ bijects $\operatorname{dom}(f)\to \operatorname{dom}(g)$.
Let $\mathcal{S}(f,g)=h$ denote the Schroder-Bernstein function just described, so now we just need to construct some injections.
For any ordinal $\alpha$, let $\mathcal{W}_\alpha$ denote the set of rank-respecting non-strict wellorders of $V_\alpha$, and let $\mathcal{C}_\alpha$ denote the set of choice functions with domain $V_\alpha\setminus\{\emptyset\}$. Notice that each $C\in\mathcal{C}_{\alpha+\omega}$ can be restricted to some $c=C\restriction_{V_\alpha}$, where $c\in \mathcal{C}_\alpha$. This method induces a partition of $\mathcal{C}_{\alpha+\omega}$ based on which functions have the same restrictions. A similar phenomenon happens with $\mathcal{W}$, so we consider the following definitions of those equivalence classes.
$$\begin{align}
[c]&:=\{C\in\mathcal{C}_{\alpha+\omega} : C\restriction_{V_\alpha}=c\} \\
[w]&:=\{W\in\mathcal{W}_{\alpha+\omega} : W\restriction_{V_\alpha}=w\}
\end{align}$$
We first construct $F_{\alpha,w,c}$ injecting $[w]\to [c]$. To do this, for each $W\in [w]$ we define $F_{\alpha,w,c}(W)$ by first taking the $\min$ function of $W$, restricting its domain to $V_{\alpha+\omega}\setminus V_\alpha$, and then extending that by $c$ to get a choice function with domain $V_{\alpha+\omega}\setminus\{0\}$. Where $C=F_{\alpha,w,c}(W)$, we get $C\in \mathcal{C}_{\alpha+\omega}$ simply because $C$ extends $c$. This $F_{\alpha,w,c}$ function is injective, demonstrable by recovering $W$ from $C$. Since $W$ extends $w$, we need only consider $a,b\in V_{\alpha+\omega}\setminus V_\alpha$. We recover the $W$-order of $a,b$ by consequence of $W$ being rank-respecting, proven by letting $\beta=\max(\operatorname{rank}(a),\operatorname{rank}(b))$ and observing the following.
$$\begin{align}
(a,b)\in W &\iff a=\min{}_W(a,b) \\
&\iff a=\min{}_W(\{a,b\}\cup V_{\beta+2}\setminus V_{\beta+1}) \\
&\iff a=C(\{a,b\}\cup V_{\beta+2}\setminus V_{\beta+1})
\end{align}$$
With more difficulty, we now construct $G_{\alpha,w,c}$ injecting $[c]\to[w]$. Fix any $C\in [c]$, so that $C$ is a choice function on $V_{\alpha+\omega}\setminus\{0\}$ extending $c$. Since $\alpha+\omega$ is a limit ordinal, an easy corollary of the wellordering theorem will construct an explicit parameter-definable function $g_{\alpha,w,c}$ sending $[c]\to [w]$. That function isn't injective, but it can be used to construct $G$. We'll assume we're using Kuratowski's ordered pair, which obeys $\operatorname{rank}((x,y))=\max(\operatorname{rank}(x),\operatorname{rank}(y))+2$ in full generality, hence any pair $(S,x)$ having $x\in S$ will obey $\operatorname{rank}((S,x))=\operatorname{rank}(S)+2$.
For each $S\in\operatorname{dom}(C\setminus c)$, we can let $q_S=(S,C(x))$, and find $p_S=\min_g\{(S,x) : x\in S\}$. Finally, define the wellorder $G_{\alpha,w,c}(C)$ as a modification of $g_{\alpha,w,c}(C)$ formed by permuting $p_S$ with $q_S$, for every $S\in\operatorname{dom}(C\setminus c)$. These permutations don't conflict, so we can apply all of them simultaneously to obtain a well-defined wellorder. We only permute terms of equivalent rank, so the resulting wellorder is still rank-respecting. Finally, we only permute terms with rank at least $\alpha$, so the resulting wellorder is still an extension of $w$. In other words, $G_{\alpha,w,c}(C)$ is a member of $[w]$ as required. This $G_{\alpha,w,c}$ is injective since, for any $S\in\operatorname{dom}(C\setminus c)$, the following holds by consequence of our permuting $p_S$ with $q_S$.
$$(S,C(S))=\min{}_G\{(S,x) : x\in S\}$$
Note that everything we've done so far works even if $\alpha=0$. To complete this segment of the proof, let $H_{\alpha,w,c}=\mathcal{S}(F_{\alpha,w,c},G_{\alpha,w,c})$, so that $H_{\alpha,w,c}$ is a bijection $[w]\to[c]$. Notice that $[w]$ is trivially in bijection with the set of rank-respecting wellorders on the domain $V_{\alpha+\omega}\setminus V_\alpha$, where the bijection just takes each $W\in[w]$ and deletes the initial segment $w$. A similar phenomenon holds for $[c]$, so the set of wellorders with domain $V_{\alpha+\omega}\setminus V_\alpha$ are in bijection with the set of choice functions on the same domain. This gives a bijection between the factors, as we needed, but now we've got these extra $w,c$ parameters to worry about. Fortunately, this won't affect the strategy too much.
The bijection between hyperclasses
Define, for each non-successor $\lambda$, a bijection $B_\lambda:\mathcal{W}_\lambda \to \mathcal{C}_\lambda$. This is done via transfinite recursion on $\lambda$, using the following rules.
$$\begin{align}
B_0(\emptyset) &:= \emptyset &\\
B_{\lambda+\omega} &:= \bigcup\{H_{\lambda,w,B_\lambda(w)} : w\in\mathcal{W}_\lambda\} \\
B_{\mu}(W) &:= \bigcup\{B_\lambda(W\restriction_{V_\lambda}) : \lambda<\mu\} \text{ whenever $\mu$ is a limit of limits.}
\end{align}$$
We can verify that each $B_\lambda$ is a bijection on the required domain, and moreover they're compatible with restrictions in the sense that $B_\lambda(W)\restriction_{V_\mu} = B_\mu(W\restriction_\mu)$ for $\lambda<\mu$. This works trivially in the base case $\lambda=0$, since $V_0=\emptyset$ and thus $\mathcal{W}_0=\mathcal{C}_0=\{\emptyset\}$.
Assuming $B_\lambda$ is a bijection, we have $B_{\lambda+\omega}$ being the union of all the $H_{\lambda,w,B_\lambda(w)}$ bijections. These all have disjoint domains, so $B_{\lambda+\omega}$ is at least an injective function. That $B_{\lambda+\omega}$ is bijective onto $\mathcal{C}_{\lambda+\omega}$ holds since the codomain is the union of the $[B_\lambda(w)]$ equivalence classes, and bijectivity of $B_\lambda$ onto $\mathcal{C}_\lambda$ guarantees that the aforementioned union is exactly $\mathcal{C}_{\lambda+\omega}$. The restriction-compatability condition is preserved during the successor stage, since we just have $B_{\lambda+\omega}(W) = H_{\lambda,w,c}(W)$ where $w=W\restriction_{V_\lambda}$ and $c=B_\lambda(c)$, and therefore $B_{\lambda+\omega}(W)\restriction_{V_\lambda} = c = B_\lambda(W\restriction_{V_\lambda})$.
Finally, the restriction-compatibility condition guarantees that $B_\mu$ is a well-defined bijection during the limit stages of the recursion, and compatibility is also clearly preserved at the limit step. That $B_\mu$ is bijective follows since we can construct an explicit inverse, having almost identical structure, as below.
$$B_\mu^{-1}(C) = \bigcup\{B_\lambda^{-1}(C\restriction_{V_\lambda}) : \lambda<\mu\}$$
The logic behind limit stages extends directly to proper classes. Given any global wellordering relation $<$, we can define a global choice function $\mathfrak{c}$ as follows.
$$\mathfrak{c}(S)=B_\lambda(\{(x,y)\in V_\lambda : x<y\})(S) \hspace{12mm}\text{where $\lambda>\operatorname{rank}(S)$ is a limit}$$
This gives an interpretation of Global Choice. Conversely, Global Choice interprets a Global Wellorder by defining the relation $<$ as follows.
$$x<y \iff (x,y)\in B_\lambda^{-1}(\mathfrak{c}\restriction_{V_\lambda}) \hspace{12mm} \text{where $\lambda>\operatorname{rank}(x,y)$ is a limit.}$$
It's seen that these two interpretations invert each other, and so we have a bi-interpretation.
