Any advice where I appear to be going wrong would be great.
It is the whole idea of using a potential divider to power a device, especially one who's current demand changes. Just don't do it because the load will make the voltage change.
If you have a 12V pump then just feed it 12V from a 12V regulator, don't forget to include the capacitors on the input and output.
I am designing a circuit to run a 12v pump, it pulls 200mA on start up then drops to about 110mA.
As you've indirectly calculated, your pump has a resistance of 60 Ohms.
If you make a regular-old voltage divider, that 60 Ohms is in parallel with your 60 Ohm resistor. That gives you a net 30 Ohms in series with the 30 Ohm resistor, messing-up your voltage divider calculations, and now you're dividing 18V in half and you've only got 9V.
Then after start-up, the resistance goes up and the voltage goes-up (I didn't bother making the calculations). But the net resistance is still less than 60 Ohms, so you never get up to 12V.
That's one of the major problems with using a voltage divider in power applications. When the load changes, the voltage changes.
The other major problem is power loss in the resistors.
Now... You could eliminate the 60 Ohm resistor and use the 60 Ohm pump in place of the resistor. That would give you the correct voltage at start-up (assuming the 200mA is an accurate spec under actual conditions), but the voltage would go up after the motor starts-up. You could calculate the resistance after start-up and change the other resistor to get the correct 12V run-voltage, but then you wouldn't have the full 12V at start-up, and under load it might never start.
And then you have lots of power loss - the zener burns up 50% of the power, so your total current draw on the 18v supply is 50% higher than it needs to be.
A Pololu D24V3F12 would work at the stated voltage and current. It's a little close to the bone though. I would step up to the next size up for only $2 more. that should handle the required power easily and efficiently. It will be about 90-95% efficient on the power conversion so you will draw less than 100mA from the supply.
