Electronic – For an Op-amp with a single power supply, how do you get negative voltage

This should work, but usually there's also a resistor from the non-inverting input to ground:

If \$\dfrac{R4}{R2} = \dfrac{R3}{R1}\$ then:

\$V_{OUT} = \dfrac{R_{3}}{R_{1}} \times (V_{IN+} - V_{IN-})\$

To minimize offset error both inputs have to see the same impedance, and therefore \$R1 = R2\$ and \$R3 = R4\$.
Omitting R4 will only give a scaling factor for the voltage on the non-inverting input, namely:

\$V_{OUT} = 2 \times V_{IN+} - V_{IN-}\$

but changing the setting for the lower potmeter should definitely have a result in the output. Did you measure the voltages on both inputs?

What happens if you set the lower voltage to 2.5V and the upper to 1V? The inverting input should also be 2.5V, and the output 4V. What do you measure?

Note: especially the lower voltage follower is not necessary in your version; the opamp's input current is low enough to be negligible for most uses, and by the way, you're connecting the potmeter's wiper to an exactly same input!

Further reading
differential amplifier tutorial (interesting site overall!)

You can generate a negative voltage with a small current quite simply with a source of AC, and a voltage doubler rectifier.

(This shows a doubler, reverse the diodes and capacitors for an inverter)

See here for some circuit diagrams.

The AC would come from your Arduino, by toggling a single pin regularly. You must add a series resistor too. It might be possible to use the PWM outputs, set them for 50% (AnalogWrite) and the pin will change state continuously.

Notes: as you are producing a square wave, not a pure sine wave, you will need a resistor to limit the current, as though the uC was driving a short circuit. If you only have 5 V, you should use Schottky diodes to get the most possible voltage out. Choose appropriate capacitor values for the frequency of AC that you have. Higher frequency is better, until you get to the point that the junction capacitance of the diodes becomes relevant. Available current will be small, perhaps 1/2 or 1/4 of the maximum output current of the Arduino, be careful not to overload the output pin, even during start-up when the C2 output capacitor is not charged.