Since the well-ordering of the classes would not be a class, but a meta-class, you'd have to say more about what you mean exactly. For example, do you want to go to third-order set theory? Or do you want to augment the structure with an additional relation that is a well order? Or perhaps one is simply interested in the case that there is a definable well ordering in this model of its classes.

But let me answer for the case of a definable well order of the classes. 

First, it is consistent that indeed there is a definable well ordering of the classes in this model. For example, if you had started in a model of V=L, then $V_\kappa=L_\kappa$, and the subsets are those that arise in the constructible hierarchy before $L_{\kappa^+}$. But all those sets be interpreted using class codes (much like reals can interpret hereditarily countable sets). In this case, you can define the $L$-order for subsets of $V_\kappa$, and there will be a (second-order) definable well ordering. 

It isn't always true in KM that you have a definable well-ordering. The reason is that KM is known not to prove the class choice principle CC, which asserts that whenever every $x$ has a class $A$ with $\varphi(x,A)$, then there is a class $U\subset V\times V$ such that $\varphi(x,U_x)$ for every $x$, where $U_x$ is the section above $x$. If there were a definable well order of the classes, then we could define $U$ to use the least class in each section, and so we'd get CC as a consequence of class comprehension. 

The CC principle is extremely useful in class theory, enabling many bi-interpretation results. 

Finally, let me point out that your models $V_{\kappa+1}$ do not always have a definable well ordering of the classes. To see this, start with $\kappa$ inaccessible and then force to add $\kappa^+$ many Cohen subsets to $\kappa$, which preserves inaccessibility. Now, with any parameter, there will be many mutually generic sets over those parameters, but in a symmetric-model type argument, there can be no condition forcing that a particular one of them is least.  So there will be no definable well order there.