I am a bit confused with Laplace domain and its equivalent time domain conversion
Consider the s-domain of first order LPF filter which is $$\frac{V_o(s)}{V_i(s)}=\frac{1}{1+sRC}$$
Now for a second order LPF filter in s-domain is simply the multiplication of the transfer function by itself i.e $$\frac{V_o(s)}{V_i(s)}=\frac{1}{(1+sRC)^2}$$ The implmentation of such a transfer function with resistor and capacitor are two RC filters cascaded like shown in the figure
Now for analysis of the above implemented filter in time domain, considering a step input the analysis of this filter is \$V_1(t)/V_{\text{in}}(t)=1-e^{-(t/RC)}\$ and \$V_o(t)/V_1(t)=1-e^{-(t/RC)}\$, and hence \$V_o(t)/V_{\text{in}}(t)=(1-e^{-(t/RC)})^2\$
But in time domain the multiplication of Laplace domain transfer function should be convolution, yet the second order RC filters are implemented as multiplications. Also the Laplace transform of \$V_o(t)/V_{\text{in}}(t)=(1-e^{-(t/RC)})^2\$ is not \$V_o(s)/V_i(s)=1/(1+sRC)^2\$
What am I missing here?
Here is an exercise I tried. Assuming \$V_i(t)=u(t)\$, unit step function, which in the \$s\$ domain is \$1/s\$, the Laplace transfer function for a first order LPF is \$V_o(s)=V_i(s)\times 1/(1+sRC) = V_o(s)=1/s(1+sRC)\$. The inverse Laplace of this function is \$V_o(t)=u(t)\times (1-e^{-t/RC})\$. This checks out which I verified in Matlab in the time and \$s\$ domains.
Now for the second order LPF and step input with a buffer in between like in the circuit by MatteoRM: the Laplace transform \$ V_o(s)=1/s(1+sRC)^2\$ right? If I follow the same exercise as before, the inverse Laplace is \$1 - (te^{(-t/(RC))})/RC) - e^{(-t/(RC))}\$. Now this does not check out in time domain. Again, what am I doing wrong?