Voltage Drops

Pulse counter has a high impedance input (probably). So, there will be practically no current through the signal wire. Therefore there will be no voltage drop across the signal wire.

I'm assuming that you're planning to make a 4-terminal system (power, GND, irrigation signal, potable signal). At the house, your signal will be measured with respect to, GND. Voltage drop across the GND wire will be added to the signal (as a DC offset). Power supply current is running through GND wire. We can't assume that this current will be negligible. You should measure it. Resistance of 18AWG wire is 6.4 Ohm per 1000ft (from tables). If the circuitry at the end of 200ft wire consumes 100mA (for example), it will cause a 0.128V drop in the GND line.

Your pulse signal is a pseudo-digital voltage signal. To an extent, it's tolerant to voltage drops across the signal and ground wires. As long as the pulse signal is greater than a threshold, the counter will count it.

Interference

For a 200ft line interference can become a serious problem. Usually, the strongest source of interference is 60Hz from AC power lines. If you hang your cable it along the power lines, it would be the worst case. Cable with properly grounding shield helps. Some signaling methods are less susceptible: 4-20mA current loop, differential signaling. For the time being, I will not go into more details here.

Lightning Protection

Gas Discharge Tube (GDT) or Spark Gaps.

Optoisolator makes sense. Keep in mind that power supply and GND lines are also susceptible to lightning. To isolate the the exposed part of your circuit completely, use an isolated power supply.

First - don't power a pump from USB unless it will consume less than 200mA. Keep that in mind for future applications/projects. DC motors will draw very large start-up currents, and this can blow your USB hubs or their fuses, and other nasty things. Of course, a 1.5A USB phone charger plugged into the wall would be fine for most things! Always be careful about your choice of power inputs, often noobies will forget or downplay the importance of checking that all of their rated inputs/output power requirements in the project are going to work.

According to a dodgy datasheet for that product I just found, it can pull up to 360mA at 7.2V, so make sure it's not running off a computer USB port.

Because the load is only 360mA though, you can get away with a nice and small MOSFET as a switch instead of a relay. Because the pump is just a DC motor, you can switch it with a low-side N-channel MOSFET configuration.

Here is an example circuit for you:

^{simulate this circuit – Schematic created using CircuitLab}

The way this works is the USB 5V+ goes into the pump pin, and the pump's other pin goes to the MOSFET's Drain pin. The MOSFET's Source pin goes to GROUND.

Make sure to tie the two difference supply's GROUND pins together.

The SIGNAL input to the MOSFET's Gate pin comes from your Arduino. When the SIGNAL goes high, the MOSFET will turn on, and act like the relay did. When the SIGNAL goes low, it will turn off the MOSFET. The pull down resistor R2 is there to ensure it actually turns off, and will also prevent a noisy start-up condition triggering the pump.

The resistor R1 is optional, but good practice.

You should also probably have a Schottky diode from the -ve Pin of the pump to the +ve Pin of the pump in the way I've drawn it, as an inductive voltage spike clamp for when the pump turns off.

I suggest you DO indeed put a 1000mF or more capacitor (go for a 16V or 35V rated one just for the funz. If you do a 6.3V one, it may pop if you don't put the over voltage clamp on it as I mentioned).

This capacitor will assist with the start-up surge currents which can cause brown-outs and other issues as you have encountered.

You may want to properly filter/protect the sensor too, by putting a 0.1uF (100nF) capacitor on the power supply leads for it.