Calculating feedback factor \$V_f/I_o\$
As the circuit is drawn, there are two contributions to the voltage \$V_f\$:

\$V_f = I_f \cdot Z_{11} + I_o \cdot Z_{12}\$

Thus, to find \$Z_{12}\$, you must zero \$I_f\$.


  When you turn off \$I_f\$, what actually happens?


What actually happens is that port 1 is open-circuited.

Conceptually, a voltmeter is connected to port 1 and a current is driven into port 2.  The measured voltage is then just the product of that current and \$Z_{12}\$.

For this circuit, the calculation is almost trivial and \$Z_{12}\$ can be written by inspection.




  Do you mind showing how to calculate this?


Find the current through \$R_2\$ (think current division) and then, using Ohm's Law, find the voltage across.  This voltage is just \$V_f\$ due to \$I_o\$.  If you set \$I_o = 1A\$, then the numeric value of \$V_f\$ equals the numeric value of \$Z_{12}\$