Finding the resistance to charge/discharge for a LIFeMP battery
In the datasheet you have a parameter called &quot;DC Internal Resistance (max).&quot; For the particular battery you are looking at (U24-12XP) this value is \$R_{DC}=6m\Omega\$.
One of the models used to represent a battery looks as follows:



simulate this circuit &ndash; Schematic created using CircuitLab
Where the diodes are ideal. Back to your question, a very detailed battery model includes two resistances, one when you are charging and one when you are discharging a battery. They simply represent loss. All that is being done is finding the best approximation to a battery behavior. There is even a more complex model that includes hysteresis voltage in series with the voltage source.
It's often assumed that \$R_{discharge}\approx R_{charge}\$ and they are in fact really close. Then
$$R_{discharge}\approx R_{charge}=6m\Omega$$
That's why you can then use the widely known battery model of the voltage source in series with a single resistor, which would be \$R_{series}=6m\Omega\$.
EDIT: In response to your comment, here's what I've got.
What I have seen, at least in the models I've worked with, is that the voltage source in the circuit above is the one indexed by the state of charge (SOC), something like\$V(soc)\$. Anyway, if you look at your graphs, there are curves for charging and discharging. Here they are:


Those plots are not necessarily the specific ones for the module your are using, but since they are in the datasheet for all of those, they should behave in a similar manner. What you see here is that the open circuit voltage (the one in the schematic above before the resistors) is dependent on the state of charge. There is no a clear way to extract \$R_{charge}\$ and \$R_{discharge}\$ other than using the one value given by the datasheet, for your module, \$6m\Omega\$. At least that's what I would go with. Make that value a constant.
Now what really does change with both the SOC and the temperature is \$V(soc)\$. If you take a look at that last two graphs,  one is for charging and the other for discharging. The various plots that you see are within the same graph are for different charge/discharge rates (C-rate).
Your first graph (charging) is for a fixed temperature of \$23^o C\$ and they try different C rates. The second graph (discharge) tried with differents temperature, but the kept the C rate constant at C/2 (\$C/2=55\$ Amps for the module you are working with, U24-12XP).
So, you would need some type of lookup table where you store all the data points you see on the graphs so that for a specific C-rate and temperature, you can then find the \$V(soc)\$. Seems to be like a tedious work, but that's how I would do that.
Answering a couple of the questions in your comment.The capacity percent on the graphs is the state of charge (SOC). The C-rate is how much current you are either charging or discharging with. For your case, U24-12XP has a capacity of 110AH, so 1C is 110 Amps, C/2 is 55 Amps, and so on. So, for example, if I told you, try a discharge rate of 5C for your battery, I mean 550 Amps.