The ideal probability of obtaining some measurement result $i$ (with associated measurement operator $M_i$) from some state $\rho$ is $Tr(M_i \rho) = p_i$. For the computational basis, $M_0 = |0\rangle\langle 0|$ and $M_1=|1\rangle\langle 1|$.
If you can only measure in the computational basis, consider trying to tell whether you have the state $\rho_0 = |+\rangle\langle+|$ or $\rho_1 = \frac{1}{2} I$. Let $p_{ij}$ denote the probability of outcome $i$ in state $j$. It is easy to show that $p_{00} = p_{01} = p_{10} = p_{11} = \frac{1}{2}$, which is to say that $\rho_1$ and $\rho_2$ are functionally equivalent as far as you can tell. If you cannot distinguish a superposition (quantum) from a mixed state (classical), then your quantum states are not any more useful to you than classical states that happen to have the same outcome probabilities w.r.t. your measurement system.
Instead, if you can measure in the $x$-basis (so $M_0 = |+\rangle\langle +|$ and $M_1 = |-\rangle\langle -|$), you would find that $p_{00} = 1, p_{10} = 0$ and $p_{01} = p_{11} = \frac{1}{2}$. By measuring in other bases, we can properly distinguish superpositions from mixed states. Another way to think about it is that quantum states are points lying on/in the Bloch sphere. If we want to distinguish one point from another, we need to know all three coordinates, which correspond to the expectation values of the operators $Z,X,Y$, each of which define a two-element POVM.
The measurements we use when dealing with the Bloch sphere are typically projector-valued measurements, called PVMs. You can think of a PVM as picking some pure state on the surface of the Bloch sphere and defining it to be "outcome 0", and the state antiparallel to it will be "outcome 1". PVMs are a special case of positive operator-valued measures (aka POVMs) which are even more general, since they need only refer to a set of positive semidefinite matrices that sum to the identity. Here are a few previous answers that might be useful as well:
