@susannasaunders 

I can help you with some of the fundamental physics.

Heat loss has a linear dependence on temperature difference according to the Heat Transfer Coefficient (HTC) for your house. Assuming that IAT is the Indoor Air Temperature and OAT the Outdoor Air Temperature, then:

Heat Loss = HTC x (IAT - OAT)

Your heat loss worked out at 4200W for a temperature difference of 21.2C, so HTC is 198.1 Watts per degree Kelvin or Centigrade (W/K) for your house. You can lower HTC by improving insulation etc.

If you want the heat loss for a different temperature difference, just multiply HTC, i.e. 198.1 by that new temperature difference. So for an IAT of 23C and and OAT of 0C, -5C, -10C, -15C, and -20C, the heat loss will be 4556W, 5547W, 6537W, 7528W, and 8518W respectively (equivalent to adding 198W for every extra degree of temperature difference).

What I'm fundamentally trying to understand is the available 'headroom' for say a 5kW ASHP for a house with a heat loss of say 4.5 kW @ 0C? 

There is almost no headroom with a 5kW heat pump and 4.5kW heat loss at 0C. Assuming that was calculated for IAT of 21C, you have 500 / 198 = 2.5C to play with. So you can have +23C indoors and -0.5C outdoors and that's it! However, that would also have the heat pump running flat out all of the time, with nothing spare for your hot water! 

Radiator output depends on the temperature difference delta-T (DT) between the room e.g. 20C and the average temperature of the radiator. Assuming a temperature drop of 10C from flow to return of the radiator, the table below shows the multiplying factor to use to find radiator outputs. For example at DT30 degrees, radiators output 51% of the quoted power at DT50. At a flow temp of 40C, the output is just 21%. However, as most heat pumps work on a flow minus return difference of just 5C, you really need to look at the factor where the radiator temperature is 37.5C, which is 0.26 or 26%. Gas boilers typically work at flow minus return of 20C, so 80C flow is needed to get a 70C average radiator temperature. You can work out radiator power output by taking the quoted value at DT50 and multiplying it by (new DT value / 50) to the power 1.3. That's all the multiplicative factor are in the table.

This is contradictory experience to what the Heatpunk model is predicting.

The reality is that with likely no insulation between your downstairs ceilings and upstairs floors and an open door policy, a lot of the heat in your house is drifting upstairs. We also keep all the doors open and upstairs typically ends up 1 degree or slightly more warmer than downstairs, when the opposite would be ideal.

This table shows how Vaillant Aerotherm heat pump efficiency drops with flow temperature. It also shows how the maximum output drops with outdoor temperature.

There is a double cost hit in coping with lower outdoor temperatures. The efficiency of a heat pump depends on the difference between the outside temperature and the flow temperature required. Lower outdoor temperatures increase that difference directly and also via the higher flow temperature needed to compensate for the higher heat loss.

The point that you are getting at, effectively a sensitivity analysis for both cost and performance capability with decreasing minimum temperatures is a really good one. I'll give it some thought and see if I can come back with at least a partial answer.