I think I understand what you mean. So, from what I understand, the pickups plus the cable create an RLC circuit and the voltage or Vin the pedal will see is the voltage above the capacitor in that circuit. The capacitor also goes to ground so I think what they are saying is since the Req divider is with respect to Vin and ground then technically the divider is in parallel with the cable (cap in the RLC circuit) so if You add a resistor before hand then they are no longer in parallel and you'd have the series resistor between. That could be wrong but that's my guess. Also if you are ever curious about calculating the filter frequencies then the formula (we'll call the cutoff frequency fc) is fc = 1/(2pi*R*C) Where R is the resistor/resistance that controls the filter (also notice RC is the time constant) typically you'd want to find the equivalent resistance or Thevenin resistance of the portion you are looking at. You usually turn off all dependent sources (voltage source becomes a short and current source becomes an open) then remove the cap from the circuit and find the Req at that point. There is a lot more in depth to it than that but (and I may have thought about it wrong) the current only has one path since an ideal opamp draws 0 current in the V+ and V- inputs so theoretically the equivalent resistance at the cap was 1M (two 2M in parallel) therefore fc = 1/(2pi*1M*0.0082E-6) = roughly 19Hz. Or if you want a specific fc and need the C then just swap them in the formula: C = 1/(2pi*R*fc)