The forward voltage is the voltage drop across the diode if the voltage at the anode is more positive than the voltage at the cathode (if you connect + to the anode).
You will be using this value to calculate the power dissipation of the diode and the voltage after the diode.
The reverse voltage is the voltage drop across the diode if the voltage at the cathode is more positive than the voltage at the anode (if you connect + to the cathode).
This is usually much higher than the forward voltage. As with forward voltage, a current will flow if the connected voltage exceeds this value. This is called a "breakdown". Common diodes are usually destroyed but with Z and Zener diodes this effect is used deliberately.
The simulation result isn't totally unrealistic. You didn't show what happens in the transient after the switch is opened, but it should be something like this:
When the switch is open, for a very brief time, current continues to flow "down" through the inductor. It would be trying to flow "up" through the diode arm of the circuit.
Now understanding the diode behavior depends on knowing about a couple of behaviors beyond just that the diode allows current flow in only one direction.
First, the diode has some capacitance associated with it. This capacitance is effectively in parallel with the "ideal diode" whose behavior we normally consider. The value of this parasitic capacitor depends on the diode bias point. This capacitance allows current to flow through the diode in reverse for a brief time, while a large reverse voltage builds up across the diode.
Second, the diode has what's called a "reverse recovery" time. While there are still carriers (electrons and holes) in the pn junction generated during the time current was flowing forward, switch the diode quickly into reverse bias can cause these carriers to flow backwards, and carry a reverse current. But this current only lasts a brief time, until the carriers are swept out of the junction.
Third, after the reverse recovery behavior ends, and the diode capacitance builds up a large reverse voltage, its very likely in a real diode, that the large reverse voltage causes electrical breakdown, which will destroy the diode.
The next key is that the simulator model very probably includes the reverse recovery behavior and the parallel capacitance behavior, but not the breakdown behavior.
So what probably happened in your simulation is, the inductor did in fact continue to conduct in the forward direction for a very short time. This caused a large reverse bias to build up on the diode (because breakdown isn't modelled). This reverse bias means the "bottom" node is at a very high voltage (relative to the "top" node). This voltage causes a proportional di/dt in the inductor, eventually resulting in reversing the current direction through it.
One way to look at this is that you have created a (damped) tank circuit between the inductor and the parasitic capacitance of the diode.
But once the current starts flowing in the counterclockwise direction, the capacitance is mostly shorted out by the diode, so it might never build up enough voltage to reverse the current back in to the clockwise direction.
Meaning, the time taken for the oscillator to reverse current direction again might be much longer than the damping time constant caused by the 100 Ohm resistor, so you never see that behavior.
Best Answer
Diode D2 is forward biased by the voltage sources. With no signal applied, the resistors form a voltage divider that makes approximately \$50\, \mathrm{V}\$ on both sides of the diode.
When input is below \$+50\,\mathrm{V}\$, nothing will flow through D1 because it will be reverse biased.
When input is above \$+50\,\mathrm{V}\$, that voltage is forced to the cathode of D2, but the 100V source can still keep up with forward biasing the diode D2 and thus (almost) the same voltage will appear on output.
When input is above \$+100\,\mathrm{V}\$, D2 will close, leaving only the 100V source at the output.
Edit: For simplification, diode with zero forward voltage drop are assumed. The graphs show that everything under \$50\,\mathrm{V}\$ is capped to \$50\,\mathrm{V}\$. That is because 50V is the DC operating point of D2 and that point is maintained until D1 is able to push current into the \$100\,\mathrm{k\Omega}\$ resistor. This current will cause voltage on this resistor to rise above its ordinary \$25\,\mathrm{V}\$. But still, the voltage is low enough that some current can still pass through D2 (from one voltage source to another). Since there is no voltage drop, voltage remains the same on both resistors (it is copied from cathode to anode). The \$100\,\mathrm{V}\$ limit is caused by the fact that the 200k resistor can never get positive voltage drop, because it would require current flowing into its top electrode, but D2 will not allow this.
To put it simple: when current flows through diode, both anode and cathode are at approximately the same voltage. When current does not flow, reverse voltage can be arbitrarily high.