Is the transfer function for the given RC network right?
This exercise is a classic and you can see that applying brute-force algebra leads to a complex expression, difficult to factor. Furthermore, you can make mistake as you did. First, assign proper labels to each of the components, \$R_1\$, \$C_1\$ etc. It is important for the correct arrangements of results. If you want to solve this transfer function the hard way, transform the first stage with an equivalent Thévenin generator taken across the first capacitor. Its output impedance is \$Z_{th}(s)=\frac{1}{sC_1}||R_1\$ while the voltage is \$V_{th}(s)=\frac{\frac{1}{sC_1}}{\frac{1}{sC_1}+R_1}\$. Then, solve a resistive divider implying \$R_{th}\$, \$R_2\$ and \$C_2\$. Good luck with that!

The easiest and fastest way is to apply the fast analytical techniques or FACTs. When dealing with passive or active elements, they can't be beaten. First, consider the dc case, \$s=0\$. What is the gain when all caps are open-circuited? It is 1 in absence of a loading resistance: \$H_0=1\$.

Then, remove the capacitors while you set \$V_{in}\$ to 0 V: replace it by a short circuit and "look" into each capacitor's connecting terminals to determine the resistance you see. This resistance multiplied by the considered capacitor forms a time constant \$\tau=RC\$, the basis of the FACTs. You do that as illustrated in the below drawings and you have \$b_1=\tau_1+\tau_2\$ in a few seconds.

Then, select one of the caps and set it in its high-frequency state (a short circuit) then "look" into the terminals of the second cap. Combine this new time constant with one that you have already determined (\$\tau_1\$ for instance) and you have the second-order term: \$b_2=\tau_1\tau_{12}\$.



Once done, you can write the denominator \$D(s)=1+sb_1+s^2b_2\$ and express the transfer function as in below Mathcad sheet. The equation as it is is difficult to use, you have no insight in what is does. Apply the low-\$Q\$ approximation (look here) and factor it with two cascaded poles. You end-up with a low-entropy expression, exactly what the FACTs will always naturally lead you to.