I used to think that was true, but it's actually not quite that simple. Vf is a function of current and some diodes have rather large resistance. If you put a silicon and germanium diode in parallel, the Ge diode will conduct most of the current at low currents, but as you raise the current, the Ge diode's Vf can exceed the Si diode's Vf. At that point, the Si diode conducts the lion's share of the current. The plot below shows some measured data comparing the D2V germanium diode with the 1N914 silicon diode. Below 4.7mA, the D2V conducts most of the current. Above 4.7mA, the 1N914 conducts most of the current. When we reverse-bias the Ge/Si pair, the Ge diode will conduct essentially all of the reverse current because Ge diodes are very leaky.I think on the example on the left, the signal will only see one diode in the pair that are oriented in the same direction. I'm guessing the one with the lower Vf?