Over designed heat sink or a well designed one

Yes, thermal resistance/conductance is a two-way street, for passive conductive cooling devices, with equal speed limits posted for both directions.

OK, now for the caveats: If your heat sink is rated for forced air flow, and you have no forced air flow, then the heat flow through the heat sink will be different. Your heat sink is rated for certain conditions at both ends. If you meet those conditions (at both ends), except for reversal of temperature conditions, then you can expect equal but opposite heat flow.

Say that your 10C/W heat sink is rated 10C/W for conduction of heat from a 1W source to still air, with the contact area with air being fins. Now, you put those fins INSIDE your enclosure, in contact with still air, and you place the outside end of your heat sink in contact with a device (say, a cold plate) that will keep that end of the heat sink 10C cooler than the inside air. In that case you will get 1W of energy flow from the fins of the heat sink to the cooling device (cold plate).

You would want to pay attention to such things as: Warm air rises and cooler air falls. Air fins are most effective when hot air can rise from them and allow cooler air to come in contact with the fins. Cooling fins, on the other hand, would be more efficient when placed such that cooled air can fall away from them.

Your specified operating conditions don't make sense. The maximum voltage drop across one of these relays is 1.4V at 10A, so you can't have 45V in and 10V out. So, the maximum power dissipation in the relay itself is 14W. The temperature drop from junction to case will be 98C. You need a heatsink with a thermal resistance of (150-98)/14=3.7C/W or less.

If you really want to keep the junction temperature below 50C you must reduce the load current. Suppose you add an infinitely good heatsink, with zero thermal resistance. The thermal resistance from junction to case is 7C/W. If the ambient (and case) temperature is 25C then you can only allow a 25C increase from case to junction, so you are limited to 25/7= 3.5A of current.