@cloudapplepi - I'll have a go at trying to explain, because I think Brandon mis-interpreted your "removal of R1" suggestion as to have no connection (open-circuit), whereas I think you were intending to replace it with a straight wire (short-circuit) ....?
So I'll assume this was your thought process.
I like to use a plumbing analogy to explain electricity (it breaks down in some places but it helps to get the point across),
Try to image the electrical circuit as a central heating system, where the battery is the boiler, the wires are pipes and the resistors are radiators.
In your diagram the ultrasonic sensor represents the boiler, with the echo pin providing hot water, and the cold water from the radiators is returned to the ground pin.
When the sensor detects something, the boiler turns on to produce the hot water which is pushed out of the echo pin and through the R1 "radiator". When it gets to the pin 18 connector, it has a choice - does it go through radiator R2 and back to the ground pin to complete the circuit, or does it go into pin 18?
This choice is decided by the resistance of each pathway. If each path was of equal resistance, you would find half the water goes through radiator R2 and half would go through pin 18.
Suppose you started to close down the valve on the radiator. This restricts the flow of water by increasing the "resistance" of the radiator. The water would find it harder to go that route, so more water would go though pin 18 instead.Conversely, if you open the radiator valve, its resistance is less and more water would flow through it and less through pin 18.
So what is the resistance of Pin 18? Brandon already gave you the answer that it is very high (infinite maybe), like a very, very narrow pipe. So how much water will flow though it? Hardly any. So you can assume all of the water is flowing through Radiator R2.
Suppose we close the valve completely on Radiator R2. It now has infinite resistance and no water will flow. No water can flow through pin18 because it is also infinite resistance. So no water can flow through Radiator R1 either, because it has no where to flow to. So what happens?
Well, the boiler is still on and trying to push water out of the echo pin and pressure is building up in your pipework.
How much pressure is pin 18 feeling? All of the boiler pressure! So you are about to kill it.
Let's try and bring this analogy back to electricity.
The flow of water is the same as the electrical current. The pressure of the water is like the voltage.
When the resistance R2 is very high, there is no current flowing in the circuit, but there is a lot of voltage (pressure) at pin 18, in fact it is the full 5v, more than the 3V3 that Pin18 can handle.
In your first post you assumed the resistor could reduce the voltage on its own. But it can't. Current needs to flow through the resistor for the voltage to drop.
It's all to do with Ohm's law that states V=IR : The voltage(V) developed across a resistor(R) is proportional to the current (I) flowing though it.
So if there is no current flowing though R1, it produces no voltage across it and the voltage at pin 18 is the voltage at the echo pin minus the voltage across R1. 5V -0V = 5V
So you need current to flow through R1 and you do that by having R2 in the circuit to allow the "water" to flow back to the "boiler"
How much current is flowing?
Well, we know from Ohm's law that V=IR. The voltage (V) at the echo pin is 5V. The resistance is (R1+R2), so the current I=V/R = 5/(R1+R2)
In your example I = 5V(330+470) = 5V/800 = 6.25mA
Now we can calculate the voltage developed across R2 alone.
So using V=IR again, V= 6.25mA * R2 = 6.25mA * 470 ohms = 2.9V
So back to your question about removing R1.
If R1 is removed and no wire replaces it, then the water from the boiler will spill onto the floor and ruin your kitchen!

In this case you don't have a complete electrical circuit and nothing will work (like Brandon's answer).
But suppose you replace it with a wire of zero resistance.
Let's go through the calculation again. How much current is flowing through the circuit now?
Using V=IR, The 5V at the echo pin = current x the resistance R2
5V= I x 470, so I=10.63mA. So more current is flowing now.
What is the voltage at pin 18?
It's the same as the voltage across R2.
So V=IR = 10.63mA x 470 ohms = 5V.
Looking at it another way, there is no resistance between the echo pin and pin 18 to lower the voltage from 5V.