Confusion in feedback for switchmode dc/dc converter
Yes, that's how the error amplifier works - the regulator tries to keep the voltage at the FB pin at Vref (1.265V in your case) through the voltage divider.
Basically to set the output to 7V, you need a divider ratio of (7V / 1.265V) - 1 = 4.53:1, so for instance 1k for the bottom resistor and 4.53k for the top resistor (the actual division is 5.53, so 7 / 5.53 = 1.265)   

If you want a tracking preregulator though, you cannot just supply a certain voltage to the FB pin to control the output voltage, as the regulator needs to be able to "find" the 1.265V again by setting it's output accordingly. You have broken the FB loop if you supply a voltage directly, and the switching will either be on or off permanently.
You need to do something like apply a scaling voltage to the bottom of the resistor divider as shown here:



To take our above 7V and the 1k:4.53k divider as an example of how it works, say we now set the bottom of the divider to -1V instead of 0V. We would now get (5.53 * (1.265V + 1V) - 1 = 11.53V at the output (we add the 1V as the bottom resistor now needs 2.265V across it for the FB pin to see 1.265V)