Why does commutativity mean that two observables can be measured together?

If two observables commute, $[hat{A},hat{B}]=0$, then this means that you can always find a common set of eigenstates. In the simplest case of the eigenvalue spectra of $hat{A}$ and $hat{B}$ being non-degenerate, then this implies that the eigenstates ${|u_nrangle}$ are the same for both: $$ hat{A}|u_{n}rangle=a_n|u_{n}rangle, hat{B}|u_{n}rangle=b_n|u_{n}rangle. $$

If you start with your initial state written in the basis of eigenstates of $hat{A}$, $|psirangle=alpha|u_irangle+beta|u_jrangle$, then if measuring $hat{A}$ you get $a_i$, your state immediately after the measurement is $|psi^{prime}rangle=|u_irangle$.

If you then want to measure $hat{B}$, you have to write your new state $|psi^{prime}rangle$ in the basis of eigenstates of $hat{B}$. Crucially, this is $|psi^{prime}rangle=|u_irangle$ because as $hat{A}$ and $hat{B}$ commute so they share the same set of eigenstates. So $|psi^{prime}rangle$ is already in an eigenstate of $hat{B}$, and when you measure $hat{B}$ you will get $b_i$ with probability 1. If you did measure $hat{A}$ again you would get $a_i$ again, and so on.

This discussion becomes more subtle when $hat{A}$ and/or $hat{B}$ have a degenerate eigenvalues spectrum, but I think the above is a good starting point to answer your question.