Why doesn't choice imply global choice (in NBG)? I thought ZFC proved the existence of an inductive well-ordering that is itself a set for any stage of V. NBG with only the regular AC should then prove/assert the existence of a class R of ordered pairs (a,b) such that either: 
a has lesser rank than b, or:
they both have rank α and (a,b) exists in the ZFC well-ordering of the complement of V(α) within V(α+1). 
I don't see that we have to quantify over classes there. By another phrasing it just unites for all V(β) their inductive well-orderings along these lines, which are already supersets of the same ordering for any previous stage. And then NBG should easily prove this well-orders V. How does this proof fail?
 A: Your first sentence is true (modulo the word "inductive"), but not in the way you mean: $ZFC$ proves the existence of many set well-orderings of each $V_\alpha$. Now, under some further assumption - say, $V=L$ - there might be a sequence of somehow canonical well-orderings of the $V_\alpha$, in which case we can indeed "glue them together" to get the global well-ordering you seek. However, if there is no such definable sequence, then we can't define your well-ordering, since we would need to pick some specific well-ordering of each of the $V_\alpha$s simultaneously - that is, find a choice function for a class-sized sequence of sets. And this, of course, is exactly global choice.
EDIT: One positive point. Even though choice doesn't imply global choice, $NBG$ is strongly conservative over $ZFC$, so in some sense $ZFC$ is not enough to show that a well-ordering of $V$ exists, but is enough to show that a well-ordering of $V$ isn't too destructive to the universe. :)
A: The other answers are great. Let me point out, however, that one doesn't need the inaccessible cardinal in Asaf's argument, because one can force over any ZFC model, and there is a pure-forcing argument, not using symmetric models (which some find confusing). Specifically, one can use the argument I provided in my answer to a previous question of Asaf's Does ZFC prove the universe is linearly ordered?. 
Theorem. Every model of ZFC has a class forcing extension that is a model of ZFC, in which there is no class global linear ordering of the universe that is definable from parameters.
By augmenting the model with its definable classes, this provides a model of GB+AC in which there is no global linear order (and hence also no global well-order). 
The proof is to perform a class product that adds two sets of Cohen subsets at each stage, and then argue that one cannot definably linearly order them in the extension, because the forcing has automorphisms that preserve any given condition, but swap the resulting sets. And so no condition can decide which order to put them in. 
A: As Noah S said, you have to use global choice to choose a well-order for each $V_\alpha$.
But we can have a concrete example for this sort of failure. Consider an inaccessible cardinal $\kappa$, and add using Easton forcing a subset for every regular cardinal below $\kappa$. Now consider the symmetric model defined by things definable by bounded parts of the forcing. Let $N$ be that model.
This can be seen as a class forcing over $V_\kappa$. The class of generics does not have a well-order by standard arguments, but each $(V_\alpha)^N$ below $(V_\kappa)^N$ does have a well-ordering. Therefore $(V_\kappa)^N\models\sf ZFC$ but nowhere in $(V_{\kappa+1})^N$ we can find a well-ordering of $(V_\kappa)^N$. In particular, $(V_{\kappa+1})^N$ as a model of $\sf NBG$ satisfies the axiom of choice, but not global choice.
