There seems to be some sort of misunderstanding here. Making a measurement of observable $A$ of a system in the state $|\psi\rangle$ does not mean we need to get a number from the calculation $A|\psi\rangle$. The issue here is that $A|\psi\rangle$ is still a vector. If you are expecting the measurement to give a value of $a$ for $A|\psi\rangle=a|\psi\rangle$ then this is still incorrect, as in general $|\psi\rangle$ will not be an eigenvector of $A$.
So how does the operator $A$ relate to the measurement of the observable associated with $A$? Well, all you have to do is express $|\psi\rangle$ in the eigenbasis of $A$ $$|\psi\rangle=\sum_nc_n|a_n\rangle$$ Quantum theory tells us that if we were to measure $A$ of our system that all we can cen determine is the probability of measuring some value $a_n$. This probability is equal to $|c_n|^2=|\langle a_n|\psi\rangle|^2$.
So, from the operator we can determine two things:
- Its eigenvalues (possible measurement outcomes)
- Its eigenvectors (what we can use as basis vectors).
And from these two things we can then determine the probability of our system to have a value of $a_n$ when we measure $A$.