Resistor Divider Theory Check
The key here is that 13 mOhm (used by the sense resistor) is very small compared to the resistances used in the divider, allowing some approximations to be made. Additionally, the key is to understand that this resistor network is not a network with equivalent resistance 12.76 mOhm. It's simply a network that will control the LM5069 in the same way that a simple 12.76 mOhm resistor would have.

To start, consider the circuit for some arbitrary current I flowing through the power bus with the sense resistor alone, and perform a conversion to an equivalent circuit. It's easiest to draw it as a Norton equivalent:







simulate this circuit &ndash; Schematic created using CircuitLab

We then perform a Norton/Thévenin conversion. We can approximate the equivalent circuit of the power bus and \$R_{SNS}\$ alone as a Thévenin equivalent with 13 mOhm Thévenin impedance, and with Thévenin equivalent voltage equal to 13 mOhm * I. Here's the circuit converted to its Thévenin equivalent, with the resistor divider attached.







simulate this circuit

Because Rth (13 mOhm) is much smaller than R1/R2, the voltage drop across it is negligible, and nearly all \$I\cdot13\text{m}\Omega\$ of voltage is applied across the divider. The divider's output voltage is approximately \$I\cdot13\text{m}\Omega\frac{R1}{R1+R2}\$, which is the same as the voltage that you would get if you had a \$13\frac{R1}{R1+R2}\$ milliohm resistor to begin with.