for both of them, there are no definitive positive/negative terminal. the Vin signal is AC in both cases. When it is positive, we'll assume that conventional current flows clockwise. When it is negative, it is in the opposite direction.
Both of these circuits are parallel circuits, with Vout parallel to RL and the diode, and then in series with the resistor R.
Since voltage in parallel circuits are the equal, when the diode is forward biased, the voltage across RL and Vout will both be at the switch-on voltage for the diode, ~0.7V (this applies to both circuits)
This is the case when Vin is positive and greater than 0.7V (i.e. diode is forward biased and conducting), and hence where Vin is greater than 0.7, Vout is capped at 0.7V
When we turn this around, the diode will be reverse biased, and that section of the parallel circuit will not conduct. Since these components are parallel, the reverse-biased diode does not affect the rest of the circuit at all, and can be ignored. what you end up with is a voltage-divider.
In circuit 1, RL is significantly larger than R (10,000 times larger), practically all of the voltage goes through it, hence the bottom peak Vout~-10V
In circuit 2, RL is equal size with R, using the voltage divider formula, we arrive at the bottom peak of Vout=-5V
hope that explains it =)
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