Separable measure space and diagonalization
Saying that a measure space is separable is not standard terminology. Here it just means that $L^2(X)$ is separable, which is obvious since it is isomorphic to $H$.
If $A$ is diagonalizable then it is unitarily equivalent to the normal operator $M_f$, so it is normal.
To see the relation with the Spectral Theorem, given $\Delta\subset\sigma(A)$ Borel, define $$ F(\Delta)=WM_{1_{f^{-1}(\Delta)}}W^*. $$ Then one checks that $F$ is a spectral measure and that $$ A=\int_{\sigma(A)}\lambda\,dF(\lambda). $$