Why does the mass-energy associated with nuclear potentials obey $m = \gamma m_0$? Here is a thought experiment that has me confused.
Imagine a particle, the nonexiston. If possible, ignore quantum mechanics and let the nonexiston be a classical particle with spooky action at a distance.
The nonexiston has a field that exerts huge forces on other nonexistons. The nonexiston field is an inverse square field. An isolated nonexiston has a very tiny rest mass $m_n$, but a pair of nonexistons separated by $r_0$ has a large associated potential of $V = M c^2$ such that their combined mass-energy is $m_0 =M + 2m_n \gg m_n$
We accelerate the pair of nonexistons from $0$ to $v$ while holding their proper length of separation constant and try to predict how much energy we will need. Without assuming a priori that field energy counts as mass, can we prove it will be $(\gamma-1)(m_0)c^2$?
Yes.

*

*The nonexiston field is inverse-square dependent so the separation potential is $V\propto r^{-1}$ dependent.

*Increasing velocity from $0$ to $v$ length-contracts the separation distance by $1/\gamma(v)$

*Therefore increasing velocity from $0$ to $v$ increases the separation energy of the nonexistons, as measured by the original rest frame, by a factor of $(1/\gamma(v))^{-1} = \gamma$.

*$V = \gamma Mc^2$

*$E = V + 2\gamma m_n c^2 = \gamma(M+2m_n)c^2 = \gamma m_0c^2$
So far so good. Relativity is internally consistent if we make the nonexiston field one for which potential goes as $1/r$.
But what if we make the nonexiston potential vary like that of the strong nuclear force?
$V \propto e^{-B r}/r$
Now we should get (for some constants $A$ and $B$)
$E = \gamma(Ae^{-B r_0/\gamma}/r_0 + 2m_n c^2)$
Which is very much not $\gamma m_0c^2$
What's going on? I feel like I must be missing something.

I realized later that I packed a variable (mass) into one of my constants (B). If we assume that that means $B' = \gamma B$ then $V = \gamma V_0$ and $m = \gamma m_0$ for $\vec r \parallel \vec v$, which is just as it should be.
 A: Separation distance is parallel to motion:
Increasing velocity from 0 to v length-contracts the separation distance by 1/γ(v)
Increasing velocity from 0 to v multiplies the force that is associated with the potential energy, which is associated with the potential, by 1.  (Because from basic special relativity we know that transformation of forces parallel to motion is: F'=F)
From those two effects we can see that the potential changes by 1/γ(v).
Separation distance is perpendicular to motion:
Increasing velocity from 0 to v length-contracts the separation distance by 1.
Increasing velocity from 0 to v multiplies the force that is associated with the potential energy which is associated with the potential, by 1/γ(v). (Because from basic special relativity we know that transformation of forces perpendicular to motion is: F'=F/γ(v) )
From those two effects we can see that the potential changes by 1/γ(v).
Notice that we did not need to know how the force depends on the distance.
A: 
Increasing velocity from 0 to v length-contracts the separation distance by $1/γ(v)$

I think your first error is here. Length contraction only takes place along the dimension parallel to motion.
Suppose your molecule of di-nonexiston is at rest and we accelerate towards it instead. Your length- contraction approach predicts that an observer moving parallel to the inter-nonexiston axis would measure a different effective mass than an observer moving at the same speed in a perpendicular direction. If your molecule is rotating, would a moving observer see its relativistic mass fluctuate?
In electromagnetism, these issues are resolved by coming up with rules for transformation of the electric and magnetic fields under boosts.
