Could you explain how this current measurement circuit works?
In a nutshell, the first opamp and the transistor work together to draw a current through the 100&Omega; resistor such that the voltage drop across it matches the voltage drop across the 0.01&Omega; sense resistor. Since the former is 10000&times; the latter, the current through the transistor will be 1/10000 of the sensed current.

The current through the transistor is then fed through the two 1000&Omega; resistors, which creates a voltage drop of

$$2000\Omega \cdot \frac{I_{LOAD}}{10000} = 0.2 I_{LOAD}$$

The second opamp is just a voltage follower (buffer) for this voltage.

HOWEVER, the circuit cannot work as shown. The two inputs to the first opamp are at a value very close to V_PWR (24V), while the power supply for that opamp is at only &plusmn;12V. No opamp can work with its inputs that far outside the supply rails. You should use one of the dedicated current-sensing chips for this part of the circuit, which are specifically designed to deal with this situation.

In order to change the range of operation from 25A to 10A, the simplest thing would be to change the 2K resistance to 5K.