Decreasing of \$V_{in}\$ in a high pass filter due to the type of capacitor used
As frequency rises, the impedance of the capacitor drops and, at a significantly higher frequency than the natural 3 dB point (2605 Hz) of the filter, you could make an argument that the capacitor becomes, in effect, a short circuit hence, the output of your signal generator will be loaded by the 13 ohm resistor.

If the signal generator's output impedance is 50 ohm, the 13 ohm resistor loading the output would reduce the amplitude down to 20% of the unloaded value. This of course adds another level of complexity to things but, providing you plot the ratio of Vout to Vin it won't make a difference.

Reducing the 4.7 uF to 100 nF means the cut-off point when Xc = 13 ohms is much higher at 122 kHz but you would see the same effect as you approached this higher frequency.

Ideally, with a 50 ohm impedance signal generator, to keep the input voltage largely constant across frequency, you should use a much higher value for the load resistor i.e. something like 1 kohm. If this were repeated with the 100 nF capacitor, the 3 dB point would be nearly the same as the original at 1591 Hz. The output would drop slightly at higher frequencies and tend to reduce to a value of about 95% of its low frequency value.