Understanding compensator's transfer function
This is all about properly writing a transfer function in a low-entropy form. In a type II compensator, you have a pole at the origin to provide infinite gain in dc (actually bounded by the op amp open-loop gain \$A_{OL}\$) plus one pole and one zero.



The original equation obtained by rearranging the division of the impedance in the feedback path with \$R_1\$ leads to \$G(s)=-\frac{1+sR_2C_1}{sR_1(C_1+C_2)(1+sR_2\frac{C_1C_2}{C_1+C_2})}\$. However, you have absolutely 0 insight in this expression. The best is to reorganize it by factoring \$sR_2C_1\$ in the numerator and highlight what is called an inverted zero:

\$G(s)=-G_0\frac{1+\frac{s_z}{s}}{1+\frac{s}{s_p}}\$ with \$G_0=\frac{R_2C_1}{R_1(C_1+C_2)}\$, \$\omega_z=\frac{1}{R_2C_1}\$ and \$\omega_z=\frac{1}{R_2\frac{C_1C_2}{C_1+C_2}}\$. In this expression, you first adjust the zero and the pole to create a phase boost at crossover and then, you adjust the mid-band gain \$G_0\$ as described in this document. The expression I gave is similar to that given in the first doc as \$G(s)=\frac{k_1}{s}\frac{1+s\tau_1}{1+s\tau_2}\$ however, you now clearly see the leading term \$G_0\$ as a gain. In page 11 of the second document, the compensator transfer function is lumped in the closed-loop equation (bottom right) so it is difficult to judge what expression they have adopted. Then in the TI document, the expression in (4) is mathematically correct and is that of the type 2 but gives absolutely 0 insight on poles and gain as it does not stick to the low-entropy format I used.