The figures in the OP's post are nicely drawn, as expected in a modern textbook, but figure 3.2(c) has the external battery reversed, which is incorrect. Here is how it works, without the needless complication of the potentiometer.
First, start with a standard Daniell cell with standard assumptions, i.e., negligible internal resistance, unimolar concentrations, and so on. Then the open circuit voltage is 1.100 V and, under light load, essentially the same. This is shown in Fig. 1 below:
As shown, electron flow is from the zinc anode to the copper cathode, via the external load resistor. Note that DMM means digital multimeter, used in voltmeter mode, and DVM means digital voltmeter. The current flow is $11 \mu A $.
Now cut the wire to the cathode and insert an external DC voltage supply that is turned on, but set to supply zero volts between its terminals. This is shown Fig. 2 below:
The external DC supply is schematically depicted as a battery supplying 0.000 V between its terminals. Its internal impedance is assumed negligible. Effectively, this 0 V battery is the same as a piece of wire: the situation is the same as in Fig. 1. The current flow is still $11 \mu A $.
Now start making it interesting. First, set the external DC supply to produce +0.500 V, as shown in Fig. 3 below:
Note how the external DC supply is connected: its positive terminal connects to the copper cathode. The Daniell cell potential is opposed by the external supply voltage and the DMM shows that the difference, which is +0.600 V, is across the load resistor. Therefore, the current flow is only $6 \mu A $.
Next, set the external DC supply to produce +1.100 V, as shown in Fig. 4 below:
No current flows because the external DC supply voltage nulls (exactly opposes) the Daniell cell voltage. Both ends of the resistor are at -1.100 V with respect to the copper electrode, so no current flows and there is no oxidation or reduction taking place in the cell reservoirs. There is no anode or cathode, just electrodes. This is the dividing line between voltaic cell operation and electrolytic cell operation.
Finally, set the external DC supply to produce +3.000 V, as shown in Fig. 5 below:
Now the electron flow is from the negative terminal of the external DC supply, through the load resistor and into the zinc electrode, where reduction will take place in that cell reservoir. This is the electrolysis mode of operation. Note the voltage across the resistor is -1.900 V, i.e., -3.000 V minus -1.100 V. So the zinc electrode is now the cathode and the copper electrode is the anode.