I think the question has become over complicated.

As a basic rule of thumb, N likes to bond three times and O likes to bond two times. In these cases the formal charge is 0. If N or O is bonded more or less than these optimal times, the formal charge is usually equal to the amount of bonds more or less than this ideal amount. Meaning that if O is only bonded once, the formal charge is -1. If O has three bonds, the formal charge is +1. We could go through and prove that but let's just take it as axiomatic right now.

Now lets consider N_{2}O. There are two ways the atoms could be arranged: N-N-O and N-O-N. Why is the former preferred?

Under the assumption the molecule has to be neutral, there are two reasonable resonance structures for N-N-O: N=N=O and N≡N-O (you could also have N-N≡O, but it will shortly become apparent why this is not typically included). The respective formal charges for each of these structures are {-1, +1, 0} and {0, +1, -1}. Assuming these structures are equally likely, there is a net 1/2 unit of negative charge on each terminal atom and a full unit of positive charge on the central atom.

For N-O-N, the reasonable resonance structures are N=O=N and N≡O-N (or equivalently N-O≡N). Here the formal charges are {-1, +2, -1} and {0, +2, -2}. On average, the two terminal atoms have a formal charge of -1 and the central atom a formal charge of +2.

In essence, to stabilize N-O-N requires a much greater concentration of formal charge on the various nuclear positions than N-N-O. This makes N-O-N unfavorable compared to N-N-O, because concentrating charge in once place is usually energetically unfavorable. (Add to this that oxygen is the more electronegative element, it is especially unfavorable for it to accommodate a +2 formal charge.)