I will now add some calculations. All calculations are in metric.
First, let me tell something about the role of water speed (flow) and of the temperature delta of a tube (output  input temp). An example on a not so warm day:
 water temperature = 20 C
 air temperature = 18 C
 the sun is shining
Stop the water in the tube, and leave it for some time in the sun, until the temperature does not rise anymore. You can do that with a garden hose. Then measure the temperature of the water in the hose. Lets say it is 30 C.
If you now flow water through the tube, the output temp will be somewhere between 20 and 30 C. If the water would flow extremely fast, the temp would be close to 20 (it would hardly heat up), and if it would flow very slow, it would be close to 30.
With a very high speed (flow), you would be taking the maximum possible energy to the pool.
If you would flow water through the tube with such a speed that the output would be 25 C, you would more or less be getting half of the maximum energy. So, it would be better to have a higher water speed (flow), so that the output water would be for example 22 C. Then you would be getting 80% of the maximum energy.
Note that this a bit simplified, I am ignoring the thermal resistance of the tube wall here.
So, the lower the delta between input and output temp, the better, and you have to compare that delta to how hot the water in the tube would get if there would be no flow.
I assumed that on good day, the standstill temp might be 4050 and I aimed for a delta of max 45 C.
The next question is, how fast will the water have to flow to get the desired delta?
Suppose that the inner diameter of a 25 mm tube is 20 mm. I used 100 m tubes.
So, 100 m of tube holds 31.4 L of water.
The surface of that tube is 25 mm x 100 m = 2.5 m2
The energy of the sun is about 1000 Watt / m2
Because the sun is shining at an angle, assume 800 W / m2
So, the tube gets 2.5 x 800 = 2000 W
The energy to heat 1 L of water by 1 C is: 4.184 kJ
To heat our tube with 31.4 L by 1 C takes 4.184 * 31.4 = 131 kJ
At 2 kW, heating the tube by 1 C takes 131 / 2 = 65.5 seconds.
To get a delta of 4 C, the water has to stay 4 x 65.5 = 260 seconds in the tube.
So, the flow must be 31.4 L in 260 seconds = 0.435 m3 / hour.
Note that this is per tube, so with 20 tubes in parallel, the total flow is 8.7 m2 / hour, which sounds like a reasonable flow with my pump.
The next question is, how much pressure loss will it cost to push that amount of water through these pipes.
I have used this website (Dutch) to calculate pressure loss:

Drukval in buisstroming
A flow of 0.435 m3 / hr through a 100 m tube of 20 mm (inner) requires a pressure of 0.123 bar.
The total length of the 50 mm feed / return tubes is about 10 m.
A flow of 8.7 m3/hr through 10 m tube of 44 mm (inner) requires 0.056 bar.
So, total pressure loss = 0.179 bar. Add some for corners, etc, and total loss = 0.2 bar.
This is reasonable in my low pressure system with an oversized cartridge filter.
0.2 bar equals the pressure of a water column of 2 m.
These calculations are not 100% exact, but it will give you an idea how to choose pipe diameters and how many pipes in parallel, etc.