Electronic – Does the electric field travel mostly on the surface of the copper in a standard circuit

A second coil probably is a good way to detect the passing of the projectile. If you have two coils wrapped around the barrel like so:

  coil 1    coil 2
==//////|===/////|===
  |     |   |    |
  +    GND  A    B

A voltage is applied across coil 1 to accelerate the projectile. This acceleration is caused by a change in the magnetic field inside that coil. Let's say it is in the rightward direction. If this is to accelerate the projectile to the right, the projectile's magnetic field must be directed left. As this projectile approaches coil 2 from the left, it causes an increasing, leftward magnetic field in coil 2. This, in turn, induces a voltage in coil 2 that will oppose the changing magnetic field. So there will be a current in coil 2 that creates a rightward magnetic field. Depending on which way you wrap the coil, you will get a positive or negative voltage across coil 2. As the projectile gets farther away, you will see the opposite voltage induced, as the magnetic field of the projectile is still directed to the left, but is decreasing in magnitude. This can be measured with a voltmeter (attach the leads to A and B) just to see the effect. If you want to use that voltage to switch the first coil off, you will need to create a circuit or use a microcontroller such as an Arduino.

Long answer short, in the above configuration, you will see a voltage induced in the secondary coil as the projectile approaches, and the opposite voltage as it gets farther away.

Farfield = 1/r. Does it suggest that potential is equal everywhere in the Farfield?

No it does not. The presence of electric field implies that there will be potential difference, in other words, no potential difference or a constant potential would imply no electric field. You are wrong in assuming \$E = \frac{V}{r}\$ for any r. Instead, $$E = -\frac{dV}{dr},$$ thus \$E\ \ \alpha\ \ \frac{1}{r}\$ would imply a logarithmic potential field (i.e. logarithmic dependence on r) rather than a constant field as you say.
If you want to know as to why \$E\ \ \alpha \frac{1}{r}\$, you can imagine a sphere of radius r centered at the source. Then, power leaving the sphere is: $$P_{rad} = (4\pi r^2)P_d,$$ where \$P_d\$ is the power density of the field. To have finite power radiated away from the source even for large r, \$P_d \ \alpha \frac{1}{r^2}\$. As, power density is related to squared of the electric field magnitude, thus we get a inverse dependence on distance for electric field. In other words, $$P_d \ \alpha E^2 \ \alpha \frac{1}{r^2} \implies E\ \ \alpha \frac{1}{r}$$

In general, power density will be of the form: $$P_d = \frac{C_1}{r^2} + \frac{C_2}{r^3} + \frac{C_3}{r^4}+...$$ For large r we just get the radiated term (first term) as the others would be negligible, but for small r's we will have small radiating term and we will have field mostly due to rest of the terms and it will constitute near field.