There are no different paths for the current unless you change the circuit.
Since your load is 1.2Ω and R2 and R3 in parallel are equivalent to 40 Ω, then most of the current is going through the load. For the time being, ignore R2 and R3. 24 V across 1.2Ω will produce 20 amperes of current. Conversely, if you have 20 A of current across a 1.2 Ω resistor, the voltage drop across that resistor should be 24 V.
What Eric suggested and I concur with is that your batteries cannot supply 20 A to that load. Higher internal resistance of the batteries than you show in your model would be a likely explanation. In other words, if your batteries really have 0.01 Ω internal resistance each, then the voltage drop due to internal resistance across the batteries in series when producing 20 A would be just 20(0.01x2) = 0.4 V and the voltage across the load would be reduced by that amount. It is quite possible that the batteries are not identical. That is, one may have a much higher internal resistance than the other. Higher internal resistance will be evident as lower voltage across the load.
So, measure the voltage drop across your load. While you are at it, measure the voltage across each battery while the load is connected.
John