we have a mathematical object called a wave function from which we can extract quantities properties or observables to describe our quantum system. To get these quantities, we will apply an operator on our wave function which is Hermitian and linear like $\hat X$ or $\hat P$ in 1D.
No, this is not the case. It is true that the wavefunction (or more generally the state vector) tells us information about our system. More specifically, it can tell us the probability of measuring the value of some observable. However, it is not the case that this information is obtained by applying operators to the state vector (unless the operator is the sum of the projection operators (unity for a complete basis), as shown below).
Instead we do the following. The state vector $|\psi\rangle$ is an abstract mathematical object. Our observables represented by Hermitian operators have real eigenvalues and permit construction of complete, orthonormal eigenbases. When we take an observable $A$ and project $|\psi\rangle$ onto its eigenbasis $\{|a_i\rangle:i=1,2,3,\dots\}$ so that
$$|\psi\rangle=\sum_i|a_i\rangle\langle a_i|\psi\rangle$$
Then the values $|\langle a_i|\psi\rangle|^2$ tell us the probability of getting a result of eigenvalue $a_i$ when we measure $A$ (assuming you normalized the above sum).
And this gets to your statement
The eigenvalues of the operator represent the possible results of carrying out a measurement of the corresponding quantity.
This is a postulate of Quantum Mechanics. If we have a Hermitian operator associated with an observable, then we can only obtain measurement results equal to eigenvalues of the operator.
The above also addresses other concerns
what does any of this have to do with linear algebra and eigenvalues?
I hope it is a little more obvious now. We are working with expressing vectors in various eigenbases.
what do these operators return? For example, if $ψ(x)=Ae^{ikx}$ then $\hat Pψ=−iℏ\frac{dψ}{dx}=ℏkψ$. $\hat P$ is supposedly the momentum operator but returns the momentum times the wave function why is that?
First, it returns the momentum times the wave function in this specific instance because you started with a momentum eigenstate already. In general you would not get just a constant times the wave function. And second, like I mentioned above, applying the operator to the state vector is not how you go about obtaining the possible momentum measurements. (I suppose this does actually apply for eigenstates since you do get back the eigenvalue, but other than that this is not the case).
Why do we forget the wave function and start using an 'eigenfunction' once a measurement is made?
This is actually a separate Quantum Mechanics Postulate. If you get some eigenvalue for a measurement then the state vector right after measurement is the eigenvector corresponding to that eigenvalue (or in the case of degeneracy the state vector is in the eigenspace corresponding to that eigenvalue).
So hopefully you see now that applying an operator on your state vector doesn't then give you the observable corresponding to system (in general your system is not in an eigenstate of an observable, so you would not expect to get just some constant back out).
But what are operators useful for then? One use is described above and is probably the most important part. But there are instances where actually applying operators to states is useful. One such example is expected values. Let's say I wanted to know the expected value of measurements of $A$. Well according to the mathematics of probability this would be equal to
\begin{align}
\langle A\rangle &= \sum_i|\langle a_i|\psi\rangle|^2a_i\\
&= \sum_i(\langle\psi|a_i\rangle\langle a_i|\psi\rangle) a_i\\
&= \sum_i\langle\psi|a_i|a_i\rangle\langle a_i|\psi\rangle\\
&= \sum_i\langle\psi|A|a_i\rangle\langle a_i|\psi\rangle\\
&= \langle\psi|A\left(\sum_i|a_i\rangle\langle a_i|\right)|\psi\rangle\\
&= \langle\psi|A|\psi\rangle
\end{align}
Another example where applying operators to the state vector is used is in the propagator of the system, which tells us how the quantum state evolves over time. We usually denote this as
$$|\Psi(t)\rangle=\hat U|\Psi(0)\rangle$$ where $\hat U=\exp(-i\hat Ht/\hbar)$ is the propagator with the Hamiltonian of the system $\hat H$.
