As you've drawn it there, it appears to me that VR1 is backwards.
An alternative is that VR1 is a negative voltage as shown in this:
Call me old fashioned, but when I see an unlabelled (as in your diagram) voltage source with polarity marked, I assume that the polarity is as shown.
I'll assume that you are using theoretical diodes with a zero forward voltage drop. In that case VR1 and VR2 represent the forward voltage drop of the diodes.
Have you calculated (preferably graphed as well) Vo vs Vi for a range of Vi (both positive and negative)?
It sounds like you have some intuitive understanding, but actually running the calculations and graphing the solution is really good to do at least once.
I would recommend (if my assumptions are correct) that you assume VR1 and VR2 are 0.7V and calculate Vo for Vi varying between -2V and +2V.
If you use the diagram you have here, you'll have some interesting results (it will look completely wrong).
The big hint is that with Vi = 0, Vo should also be 0. Is this the case with the current circuit? What is Vo for this circuit with Vi = 0 and VR1 = VR2 = 0.7V?
Once you've done this, consider that VR1 and VR2 represent the forward voltage of the diodes. How is this related to what happens on the Vi Vs Vo graph?
Now consider a practical circuit that doesn't have VR1 or VR2, but relies on the forward voltage.
Your idea of what it could be used for is reasonable.
In practice, if you want higher voltages (and this principle is used on things like the inputs to power supplies to limit voltage spikes) then a slightly different circuit is used (relying on reverse breakdown of diodes and similar devices).
In other applications two diodes can be used in a slightly different manner to clip signals to within a diode drop of the supply rails.
These latter 2 uses are far more widely seen that the one you have, however this one is good to demonstrate the idea.