Assume we have a quantum system and its wavefunction $\psi$, but the wavefunction is not an eigenfunction of some operator (e.g. the z-component of the angular momentum $\hat{L_z}$).
In this case, how are we going to measure $L_z$?
Is it always the case that we can write $\psi$ as a sum of eigenfunctions of $\hat{L_z}$, and from that we extract an $L_z$ measurement?
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 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:
And from these two things we can then determine the probability of our system to have a value of $a_n$ when we measure $A$.
We can always move to the eigenbasis given a hermitian operator like $L_z, H, L^2, \cdots.$ There is a theorem called Spectral theorem, which states that there exists an eigenbasis given hermitian operator, and this applies to finite dimensional, infinite dimensional, including Hilbert spaces.
Thus, since we have an eigenbasis, we can expand $|\psi\rangle=a_i|i\rangle$ in that basis with coefficients given by $\langle i | \psi \rangle$ where $| i \rangle$ is the given eigenbasis of the given hermitian operator $A, A|i\rangle=A_i|i\rangle$.
Then we are just left with writing down the eigenvalue problem $$\langle A \rangle = \langle \psi | A \psi \rangle = \langle i|a_i^* A a_j|j \rangle=\langle i|a_i^* a_jA_j|j\rangle=A_ja_i^*a_j\langle i|j\rangle = A_ja_i^*a_j \delta(i,j)=A_i|a_i| ^2. $$
It may not be an eigenstate of $\hat L_z$ but, if the system is in a pure state, it will be an eigenstate of $\hat L_{\hat n}=\hat n\cdot \vec L$, i.e. it will be an eigenvector of the projection of angular momentum in some direction $\hat n$.
It may be tricky to find this direction but one way might be to make a beam that travels through weak magnetic field and reorient the field gradient until one gets a single spot on a screen, i.e. until all particles are deflected in the same direction by the same amount.
Since $L_z$, together with $\vec L\cdot\vec L$ are a set of commuting hermitian operators, we are guaranteed their eigenvectors form a complete set, i.e. that any state can be expanded in these eigenvectors.
In our particular case, it’s not hard to write the eigenvectors of $\hat L_{\hat n}$ as a linear combination of eigenvectors of $\hat L_z$: just find the rotation ${\cal R}$ that takes $\hat z$ to $\hat n$ and rotate the eigenstates of $\hat L_z$ accordingly.
