inverse image with compact support In several notes, when introducing operator $f^{!}$ (as an adjoint of $f_!$, actually $Rf_!$), we have to pass to derived category. I am wondering what the reason is. (Manin and Gelfand's homological algebra book explains it briefly (say Page 229) but it does not make much sense to me, especially the equality after III.64). It will be very helpful if this can be elaborated more clearly.

I don't know what kind of reason you are expecting. Here is one : $f_!$ is not right exact, so it cannot have a right adjoint. It happens that $Rf_!$ do have a right adjoint (and a useful one).

Another reason : assume that $f_!$ has an adjoint $f^!$ and we want to compute it : we will use the chain of equality $$ f^!\mathcal{G}(U)=\operatorname{Hom}(\mathbb{Z}_U,f^!\mathcal{G})=\operatorname{Hom}(f_!\mathbb{Z}_U,\mathcal{G})$$ Now take $f:X\rightarrow pt$. In that case $f_!\mathbb{Z}_U$ is the zero sheaf... (There is no compactly supported section of $\mathbb{Z}_U$). This mean that $f^!\mathcal{G}$ is the zero sheaf. So really not interesting.