Here's how the LFO works. It's kinda long-winded, but trust me, this is the short explanation. If all you want to know is the formula for the frequency, you can skip the first two paragraphs.
IC4.1 is a schmitt trigger. It's a comparator with two thresholds, determined by the power supply, Vref_B, R19, R20 and the present state of IC4.1's output. IC4.2 is an integrator. Its output ramps up or down linearly at a rate determined by the RATE pot, R18 & C10. The transfer function of a capacitor is i = C * dv/dt. Where i is the current flowing thru the capacitor and dv/dt is ramp rate of the capacitor voltage. The current in the cap (C20) is the same as the current flowing thru the RATE pot & R18. For simplicity, we'll call R18 + RATE = R. Since the output of IC4.1 is lightly loaded, we can assume that the output swings very close to the rails (0V & +9V). Vref_B is at 4.5V. That means the voltage across R is either +4.5V (when IC4 pin 1 is at +9V) or -4.5V (when IC4 pin 1 is at +0V). The current in R is ±4.5V / R. When IC4 pin 1 is at +9V, the current is positive and the voltage at IC4 pin 7 ramps in the negative direction. When it hits the lower threshold of IC4.1, the output of IC4.1 switches to ground. Now the current in R is negative, so the voltage at IC4 pin 7 ramps in the positive direction. When it hits the upper threshold of IC4.1, the output of IC4.1 switches to +9V and the process repeats. The waveform at IC4 pin 7 is a triangle wave with an amplitude determined by IC4.1's thresholds.
IC4.1 will switch whenever pin 3 crosses 4.5V. We can calc what the voltage at IC4 pin 7 will be when IC4 pin 3 is at 4.5V. There are two solutions: one when IC4 pin 1 is at 0V and one when IC4 pin 1 is at 9V. We'll do the 0V case first. When pin 1 is at 0V and pin 3 is at 4.5V, the current thru R19 is 4.5V / 100K = 45uA. Since no current flows in or out of pin 3 (there is a tiny bit of current, but not enough to matter), the current in R20 is also 45uA. That makes the voltage across R20 45uA * 68K = 3.06V. The right-hand end of of R20 is at 4.5V and the current is flowing left-to-right, so the left-hand end of R20 is at 4.5V + 3.06V = 7.56V. Because Vref_B is half-way between the rails, the thresholds for IC4.1 are symmetrical. That means that the other switching threshold is 4.5V - 3.06V = 1.44V. The difference between the thresholds is DV = 2 * 3.06V = 6.12V.
Another way to calc DV is DV = 9V * R20 / R19.
Here's how we get a formula for frequency:
The time it takes C10 to ramp up and back down again (one cycle) is T = 2 * C10 * DV / i
The current in R is i = 4.5V / R. Sub DV and i into the equation for T.
T = 2 * C10 * 9V * R20 / R19 / (4.5V / R)
Rearranging, we get T = 4 * C10 * R20 / R19 * R
Frequency F = 1 / T = R19 / (4 * C10 * R * R20)
When the RATE pot is at min resistance, R = 47K. When the RATE pot is at max resistance, R = 1047K.
The max freq is F = 100K / (4 * 1uF * 47K * 68K) = 7.8Hz
The min freq is F = 100K / (4 * 1uF * 1047K * 68K) = 0.35Hz
So how do we change the frequency range?
Changing R19 or R20 will change the amplitude of the triangle wave, so we don't want to mess with those. We probably don't want to change out the RATE pot. The ratio of R18 to the resistance of the RATE pot sets the ratio of max to min freq. Also, we can't make R18 too small or it will load the outputs of IC4 too much at the max RATE setting. That leaves C10 as the best way to change the frequency range. There are practical limits on how big we can make C10. Electrolytic caps leak too much current to work in this circuit.