The voltage on pin 2 of the LTC1042 is the center of the window, and the voltage on pin 5 is half the width of the window. In your circuit pin 2 is set to 2V by Zener diode D1, so the center of the window is 2V.
R6 and R8 form a voltage divider with a ratio of (R6+R8)/R8 = (180k+20k)/20k = 10. The divider is powered by 2V, so its output voltage is 2V/10 = 0.2V. This is half the width, so the window is 2V+-0.2V = 1.8V to 2.2V.
Some other component choices in your circuit are perplexing.
The purpose of R11 is unclear. Pin 2 has a low but poorly defined bias current of +-0.3nA, so adding resistance in series would not predictably alter the window center. The datasheet does not show any examples of resistance being inserted between pin 2 and the window center voltage. The only useful purpose I can think of is to filter out noise, but this would require adding a capacitor to ground.
The bias voltage on pin 3 is set (via R2 and R7) to half the supply voltage, ie. 2.5V. The Thevenin equivalent resistance at this point is only 10K || 10K = 5kΩ, so a foot-strap resistance that varies from 1MΩ to 10MΩ will have little effect on the voltage. 1MΩ to ground will only pull the input voltage down to 2.488V which is well short of your window, while 10M will barely move it to 2.4988V.
simulate this circuit – Schematic created using CircuitLab
To improve sensitivity the bias resistance needs to be much higher. However values over 10KΩ start to introduce sampling error, so an input bypass capacitor or input buffer amp is recommended (see datasheet page 5). The combination of long wiring and connection to a human body will probably introduce a lot of EMI (particularly at mains and low rf frequencies) which an input bypass capacitor would help to reduce.
R3 sets the sampling frequency to the maximum possible (~10KHz). This lowers the effective input impedance and increases the error caused by high source resistance. I see no advantage in having such a high sample rate. 100Hz should be sufficient, and would increase effective input impedance to over 100MΩ.
You have stabilized the window voltages with a Zener diode, but the input bias voltage is determined by the supply voltage. This means any variation in supply voltage will affect the input voltage relative to your window. It might be better to remove the Zener and derive the window voltages directly from the supply voltage.
Alternatively you could use a TL431 'precision shunt regulator' to stabilize the bias voltage at 2.5V (with a single bias resistor from there to the input) and divide this voltage down to produce the center and width/2 voltages. Compared to a low voltage Zener the TL431 is much more accurate and has a lower temperature coefficient.
If you want to detect strap resistances within a particular range (eg. 1-10MΩ = green, <1MΩ or >10MΩ = red) then you need to calculate the input voltages that will occur and set the window parameters to match.
For example if the effective bias resistance is 500k (R2 and R7 = 1M, or a 500K resistor connected to 2.5V) then with 10M to ground you have a voltage divider which changes the input voltage to 2.5V / (10M+500K)/10M = 2.38V, while 1M will change it to 2.5V / (1M+500K)/1M = 1.67V.
simulate this circuit
To detect this range you would use two voltage dividers setting the window width/2 to (2.38V-1.67V)/2 = 0.355V and the center to 2.38V-0.355V = 2.025V. Rather than calculating resistor values and then having to find precision resistors to match them, it might be easier to just install two multiturn trimpots and adjust them to the required voltages.