Is there any relation between the input and output of bridge rectifier that can explain this sentence?
This means that the bridge will only pass current from the input to the output when the input voltage (whether positive or negative) exceeds Vbuf by two diode drops (1.5 to 2 V depending on how high the current is).

What the images try to represent (IMHO rather poorly) is that, given a sinusoidal input, the bridge will only charge Cbuf in the periods of the sinusoid where the voltage is high enough. In the other periods, when the input voltage is too low, the bridge will not conduct and the load will take it's current from Cbuf. That's why the voltage a Cbuf approximately resembles a saw tooth: going up during charge, going down during discharge.