Current sense amplifier
As with anything in engineering it is all about doing the right approximations, so as to simplify the required equations and prime our intuition. . 

You can see the circuit in several different ways, all of which would be correct to one degree or another. All of which can bring some insight on the working of the circuit. 


As a slightly modified Widlar current source, that requires solving a transcendental equation. 
As a common-base differential amplifier with \$I_FR_{SNS}\$ as its own input, \$~V_{OUT}\$ as the bias/common mode voltage, and RB and \$R_{FB2}\$ as the feedback network. 
As a common-base amplifier biased by a diode-connected transistor and With \$R_{FB2}\$ as feedback. 
As a common emitter amplifier. With \$R_{FB2}/R_{FB1}\$ as a good approximation to its gain. 
As individual transistors solving approximate KCL equations.
Use all of the above and examine the limit conditions. 


As a Widlar current source, you know that the current on the Q2 branch will define the mirror bias and, depending on \$R_{SNS}\$ and the output current, will be higher than the current through the Q1 branch. You are designing for \V_{OUT}\$, you know the sense resistor drop will be small, so you choose a reasonable bias for the mirror and choice of transistors. 1mA is a reasonable bias in many cases, so would be 0.1mA, and 10mA seems wasteful. This choice mostly affects the frequency response of the amplifier which could interact with the regulator feedback loop and affect stability. 

But here is where your analysis gets into trouble. You assume that Vbe of both transistors has to be the same, which automatically forces their currents to be the same, which also goes against what we know about that type of current mirror. 

The reason TI assumes 1.25V across \$R_{FB1}\$ With 1mA in it, is because this is the desired bias condition and that is the voltage imposed by the feedback loop of the regulator. The only way this can happen is if the drop across RFB2 and RSNS is the same. 

So the ratio \$R_{FB2}/R_{SNS}\$ multiplied by the bias current of 1mA determines the design’s output current.

So. What happens if there is no load? The output voltage will increase until 1mA goes through RFB1. This will require a large current on RB, as determined by the Widlar mirror. So RB and RFB2 determine the maximum output voltage (figuring it out requires solving the transcendental equation).

But more to the point, \$R_{B}/R_{FB1} * 1.25V + 0.7V\$ determines the output voltage at the desired operating condition. 

Do note that there is a balancing act. Both Vout and IF are being regulated by the feedback loop. But the gain for the current is \$R_{SNS}  * R_{FB2}/R_{FB1}\$ while for the voltage is \$R_B /R_{FB1}\$