Intuitive to get pole and zero points for lead compensator
Yes, there is an intuitive approach and it with the fast analytical techniques (FACTs). Reduce the excitation to 0 V (\$V_1=0\$ and look at the resistance driving the capacitor (you temporarily remove the cap and "see" what resistance is offered between its connecting terminals). You "see" \$R_1||R_2\$. This is it, you have the time constant of this circuit \$\tau_1=C_1(R_1||R_2)\$. The pole for a 1st-order circuit is the inverse of the time constant. Therefore \$\omega_p=\frac{1}{C_1(R_1||R_2)}\$. For the zero, what condition in this circuit would prevent the excitation \$V_1\$ from producing a response? In other words, what condition creates a null in \$V_2\$ despite a signal in \$V_1\$? When the parallel association of \$R_1\$ and \$C_1\$ become a transformed open-circuit. The pole of the impedance \$R_1||C_1\$ is our zero: \$\omega_z=\frac{1}{R_1C_1}\$. Then, for \$s=0\$, we have \$H_0=\frac{R_2}{R_2+R_1}\$.

Voilà! The complete transfer function is therefore:

\$H(s)=\frac{R_2}{R_2+R_1}\frac{1+sR_1C_1}{1+sC_1(R_1||R_2)}=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$

It is difficult to do simpler and faster. With some habit, you can visualize    in your head the various time constants and instantaneously infer the transfer function. Just as I did. Checkout the seminar I taught at APEC in 2016 and the examples solved in the book:

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf