We sometimes use a water pump circuit analogy to help explain circuits. It's not a perfect analogy so don't take it too far ...
Think of the battery as a water pump. Think of the diode as a non-return valve (current can only flow in the direction of the arrow). Think of the resistor as a narrow piece of pipe which restricts or resists current flow.
The way you've inserted the diode it is forward biased so current will flow. How much current depends on the resistance. If we had no resistance (big fat pipe) we would get huge flow and the pump wouldn't be able to keep the pressure up (the voltage would collapse).
It should be fairly obvious in the analogy that it won't matter whether you put the non-return valve in the supply or in the return - the current will be the same.
So, provided we don't overload the battery (resistance too low) the 5V will remain fairly constant and the current controlled by the resistance. If you were able to measure the pressure / voltage at various points along the narrow pipe / resistor you would find that it is falling from 5 at the top to 0 at the bottom.
I hope that makes sense.
One other point that may help later: if the non-return valve has a spring in it to close it when there is no flow then it will take a certain amount of pressure to crack the valve open before there is any flow. As the pressure rises the valve will open more and more until fully open. Diodes behave in a similar fashion in that they usually require about 0.5 to 0.7 V before they will conduct. In your circuit this could be measured as 0.7 V across the diode and 4.3 V across the resistor.
To everyone else, I know the analogy isn't perfect so ...