If the anode of a diode has positive voltage with respect to the cathode, why isn't the resistor/source combination modeled like this?
When the device is reverse biased, the anode does not have a more positive voltage than the cathode, it has a more negative voltage.
While the term anode technically means the electrode into which current flows, in practice it is used to mean the electrode into which current most typically flows. When a reverse bias is applied to the device, we don't rename the cathode to be the anode and the anode to be the cathode.
Even in zener diodes, which are most often used in "reverse" mode, we still call the p-doped side of the junction the anode and the n-doped side of the junction the cathode.
However, when I use a Voltage less than the 0.6V "turn-on-voltage", I get a negative current. What is this negative current telling me?
Notice that in the piecewise linear model, the model with a 0.6 V source and small resistor only applies when the applied voltage is greater than 0.6 V.
So when you get a negative current, it's telling you that you're using the wrong piece of the piecewise linear model and you should be using the piece where the current is 0 for applied voltages between the reverse breakdown voltage (-6 V in your example) and the forward turn-on voltage (+0.6 V in your example).