Consider heat, entering a single square of foil. Perhaps from the 2mm tab of SOT-23, or 1cm^2 tab of TO-220. What is the thermal resistance to the surrounding PCB. Surround that single square, of whatever size, with a 3*3 grid, the heat injected into the middle square. There are 8 squares around the center; the thermal resistance out of that center square is $$(70degree C/watt) / 8$$ or 9 degree C/watt. Now consider a 5*5 grid. What additional thermal resistance does the outer ring 4+4+4+4 squares contribute? Just divide $$(70degree C/watt)/12$$ or 6 degree C/watt. If the original IC (heat source) has Rthermal of 25 degree C junction_to_case, then we just add 25 + 9 + 6 = 40 degree C/watt.
simulate this circuit – Schematic created using CircuitLab
What if these 5*5 squares need to dump heat into the underlying GND plane? The Rthermal of epoxy-fiber glass is about 200X that of Copper foil. You can use the Rthermal of copper cubes, 1/(340 watts/degree C * meter) or 1/(3.4 watts/degree C * cm) and scale down further for 20 mils or 60 mils thickness. Then scale up by 200X, as approximation for FR-4.
Get a quadrille pad, and start sketching heat flows through grids, laterally, and vertically through vias and FR-4. Or generate 2_D grid of resistors in SPICE, with foil on the surface using low values, and FR_4 between layers using 200X higher Rvalues (in both x & y, to make a cube of FR-4).