TL;DR: The plot shows that Chang'e 4 is not locked to any external source but uses its own on-board clock to send data.
To answer this, we can ignore the Queqiao relay satellite and assume that we have a direct connection between Earth Chang'e-4. We treat Queqiao as just some kind of passive mirror for the signal.
The sinusoidal shape is produced by the Doppler effect as the satellite is in orbit around the Moon. The orbital velocity is about 1.5 km/s. This is the velocity it approaches us with when it comes visible and the velocity it moves away from us shortly before it vanishes behind the moon.
The total change in velocity is therefore $\Delta v = $ 3 km/s which converts to a Doppler shift of
$$\Delta f = \frac{\Delta v}{c} f = \frac{3}{3 \cdot 10^5} \cdot 8479.77~\rm{MHz} = 84.8~kHz$$
If we look to the frequency spectrum we notice that the peak-to-peak amplitude of the frequency shift is 85 kHz - quite precisely the calculated one.
Many space probes operate in a mode that they receive a signal from Earth and use exactly the frequency of the received signal to transmit their data back. When these don't receive a signal, they use their own, on-board clock to generate the signal. As two separate clocks are never precisely the same, the switch between both modes can be visible as a frequency jump in the spectrum.
In both modes the observed amplitude of the Doppler shift is different: We calculated the shift in the case of the "unlocked" transmitter - the spacecraft transmits always at the same frequency, but due to its motion we observe a changing frequency. In the "locked" mode, we would see twice the Doppler shift: It gets doubled because the spacecraft already receives a changing frequency signal from Earth, then uses this shifting frequency to transmit a signal that again is Doppler shifted on the way back to us. This is clearly not the case for Chang'e 4.
This "double Doppler" shift is visible for example in this recording of the Beresheet probe, again by Edgar Kaiser @df2mz. You can see how the received signal follows two separate sinusoidal curves: One with a larger amplitude for "locked" operation and a second one with half the amplitude during "unlocked" phases.
Now we assumed that Queqiao is just a "passive mirror" for the signal. There is a hint in the diagram that tells us that this is wrong: Two times, right when the signal vanishes, we see a sudden jump in the frequency (the two short, white lines). This seems to be the change between "Queqiao is locked to Chang'es signal" and "Queqiao uses its own clock" to send data.
Footnote: Thanks to @df2mz for the conversation and pointing out that I was using the wrong speed of light in my first attempt to write this.