Help me understand this voltage to frequency converter circuit
You have probably gone too far in changing resistors values becouse, even though this is going to change both duty cycle and voltage-to-frequency conversion ratio, there exists a full range of allowed values.
So let's focus on the integrator

Very first point is to rule out any common mode issue.
Infact a quick look to OPAx189 datasheet clears that input voltages can span from GND-0.1V to Vcc-2.5V.
So given the 5V power supply and the 1:2 divider R4/R8 we are allowed for a control voltage V1 in the 0V to 5V range.

The second easy point is the R4/R8 voltage divider itself.
While it should be obvious that, as far as the load on V1 is up to the mark, the resistor value is not of great concern but their ratio is instead going to affect the VFC operation.
So, for sake of generality, we define
\$\alpha=\frac{R_8}{R_8+R_4}=\frac{v_+}{V1}\$
as the ratio between opamp  non-inverting input and control voltage.

And now the kernel, assuming the opamp is within linearity we can rely on the well known approximation \$v_+=v_-\$ which, in this case, turns into \$v_+=v_-=\alpha V_1\$
The integrator input quantity is \$i_\text{in}=i_\text{R1}-i_\text{R2}\$ current.
Neglecting the T1 \$r_\text{DS(on)}\$ we can easily write it down in the two cases of T1 on and off.
\$i_\text{in(ON)}=i_\text{R1}-i_{R2}=\frac{V_1-\alpha V_1}{R_1}-\frac{\alpha V_1}{R_2}=V_1\left ( \frac{1-\alpha}{R_1}-\frac{\alpha}{R_2} \right)\$
\$i_\text{in(OFF)}=i_\text{R1}=\frac{V_1-\alpha V_1}{R_1}=V_1\frac{1-\alpha}{R_1}\$
Now given the output of U2 is the integral in time of this current the condition for having an oscillating output is this current to be sometimes positive and sometother negative.
Otherwise integrator output would continue to rise(fall) till hits the opamp power rails.
So given \$i_\text{in(OFF)}&gt;0\$ we shall find conditions for \$i_\text{in(ON)}&lt;0\$
\$i_\text{in(ON)}&lt;0\;\Rightarrow\; \frac{1-\alpha}{R_1}-\frac{\alpha}{R_2}&lt;0\;\Rightarrow\;R_2&lt;\frac{\alpha}{1-\alpha}R_1\$
So, in short, the condition for having oscillations is
$$R_2&lt;\frac{\alpha}{1-\alpha}R_1$$
Which, with your values, boils down to R2&lt;R1, i.e. any value from 0 to 2k will do.

Frequency and duty cycle relations  to be done