I have a question regarding how diode clamps work in a ideal circuit.
Is my issue that I am taking a non ideal model of the diode with a Vf of 0.6V while taking a fully ideal model of a voltage source with no output impedance to separate the two nets ?
Yes that is the problem. You need series resistance somewhere. It is not optional, ideal or not, and you often add extra series resistance to limit the current through the diodes so they don't fry the diodes or the driver, and to limit disruption from current being injected into the power rail.
Assuming you mean "zero output impedance" and assuming Vdd is actually 5.0V.
You have the equivalent of immovable object meets irresistible force. Your models are not accurate enough to describe what will happen.
In reality there is no such thing as zero output impedance for either the 6.0V or 5.0V sources. In reality the diode will not have a 0.6V voltage drop at high current.
Let's simulate what would happen assuming the voltage sources were ideal, and assuming a real diode 1N4004 or 1N4148:
simulate this circuit – Schematic created using CircuitLab
As you can see, the voltage across a real diode simply increases to meet the "immovable" ideal voltage sources. In both cases the diode current exceeds the maximum DC current rating so the diodes will fail, but for a short time they will do what is shown, fairly closely.
Adding some series resistance will decrease diode current and allow the clamping to occur without excessive current, however note that the voltage sources also may not be ideal. In particular the 5V source may not be capable of sinking current.
Here you can see that the 10V input is not clamped to 5.6V but rather the 5V output increases to 9V+. The Circuitlab LM7805 model is not accurate enough to demonstrate this behavior, by the way, so I used LTspice.
