Narrow active band pass filter design
the op-amp is correctly wired up with in an inverting circuit configuration.  because of the negative feedback through passive components, the "-" terminal is a virtual ground. the node equations (\$V_2\$ is the voltage at the node where are \$R_1\$, \$R_2\$, \$C_4\$, and \$C_3\$ are connected) are:

$$ \left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 \right)V_2  - sC_4 V_\text{out} =  \frac{1}{R_1} V_\text{in}$$

$$  sC_3 V_2 + \frac{1}{R_5} V_\text{out} = 0 $$

from that, i get

$$ \begin{align}
A(s) \triangleq \frac{V_\text{out}}{V_\text{in}} &amp; = \frac{-\frac{1}{R_1} s C_3}{\frac{1}{R_5}\left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 \right) + (sC_4)(sC_3) } \\
 \\
&amp; = \frac{-\frac{1}{R_1 C_4} s}{\frac{1}{R_5 C_3 C_4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) + \frac{C_4 + C_3}{R_5 C_3 C_4} s  + s^2 } \\
 \\
&amp; = \frac{-H \omega_0 s}{s^2 + \frac{\omega_0}{Q} s + \omega_0^2} \\
\end{align} $$

equating the corresponding coefficients...

$$ \omega_0^2 = \frac{1}{R_5 C_3 C_4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)  $$

$$ \frac{\omega_0}{Q} = \frac{C_4 + C_3}{R_5 C_3 C_4}  $$

$$ H \omega_0 = \frac{1}{R_1 C_4} $$

i think the intent, in the lecture notes posted in the question is that \$ \omega_0 \triangleq 2 \pi f_\text{m} \$

so let \$ C_3 = C_4 \triangleq C \$  and let \$ k \triangleq \omega_0 C \$.

then  $$ \frac{1}{R_1} = H \omega_0 C $$ $$ \frac{1}{R_2} = (2Q - H) \omega_0 C $$  $$ \frac{1}{R_5} = \frac{1}{2Q} \omega_0 C $$.

so plug this in for \$R_1\$, \$R_2\$, and \$R_5\$ and see if equality in the three "corresponding coefficients" equations above is met.  if so, the transfer function, as given in the question, is correct.