I think you have more serious problems than the heat issues raised by the posters so far. The following comments are based on what my tired old eyes tell me from your photos - could be wrong.

It appears that you have all eight 0.5 Ohm resistors connected in parallel, which would result in a load resistance of 0.0625 Ohms. A 48V battery system will have about 57.6V at the absorption point while charging, which is when the controller would begin to dump excess available current. This would result in a peak current of 922A. Obviously, the controller wouldn't be very happy.

This is a common problem for people not very familiar with Ohms Law, where they are thinking in terms of resistor wattage instead of the electrical circuit.

It's obvious that you need a much greater load resistance than 0.0625 Ohms to limit the current for the controller. A single 0.5 Ohm, 240W resistor has a maximum current rating of 21.9A (I=Sqr Root(P/R). Since you need a much greater resistance, you will need resistors in series to obtain it. The single resistor current rating will then determine the maximum load current (21.9A). To limit the current to this value would require a minimum resistance of 2.63 Ohms at 57.6V (R=V/I).

You can get close to this by using seven of your resistors. Connecting five of them in series plus two in parallel (the combination also in series) would give you 2.75 Ohms total. This will give you a load current of 20.9A at 57.6V and a dissipation of 1204W. The maximum dissipation of a single resistor would be 218W.