Art Of Electronics 2.10B
You kind of have the right and wrong idea. Simulation wise you can't do what you're doing because while placing a voltage supply across a component does force the voltage across that component, it also adds POWER to the circuit which is not what happens in reality.

This is the schematic you are asking about:




Now look at this:







simulate this circuit &ndash; Schematic created using CircuitLab



What is the current through \$R_3\$ here?

If \$V_{be} = 0.7V\$ then \$I_{R_3} = \frac{V_{R_3}} {R_3} = 0.7mA\$ because \$R_3\$ must have 0.7V across it.

What's the current running through \$R_2\$?

If we neglect Q2's saturation voltage (I'm just going to assume 0V here) then the voltage across R2 must be whatever is left over from the 15V supply after the above is \$V_{R_2} = 15 - 0.7V = 14.3V\$. That means the \$I_{R_2} = 4.3mA\$

Which makes sense because \$V_{be}\$ is also conducting current. How much is it conducting? The portion of 4.3mA that is not going through \$R_3\$ which is a majority of it.

I'm not sure why the book is using 0.6V for \$V_{be}\$ instead of 0.7V.

EDIT: It was proposed that 0.6V was used because the authors had a power PNP in mind rather than a small signal PNP.