When a quantum system have a double degenerescence at one point, the Hamiltonian should be proportional to Pauli matrices near this point (also known as diabolic point) [Ref.]. But, why the Hamiltonian have this form?
Dear Leandro, it's because the Pauli matrices, together with the $2\times 2$ identity matrix, form a full real basis of all the Hermitian $2\times 2$ matrices (note that $2\times 2$ Hermitian matrices depend on four real parameters), and the identity matrix is irrelevant in a Hamiltonian because it's just a conventional energy shift that acts on all vectors equally and doesn't create any subtleties such as line crossing so one may omit it in any discussion of interesting physical effects.
So any $2\times 2$ Hermitian matrix that "matters" - and yes, Hamiltonians have to be Hermitian operators - is a real linear combination of the three Pauli matrices. They don't have to be interpreted as a spin of any kind. But they're still a more natural basis than any other basis of the Hermitean matrices because the product of any pair of the matrices generates a multiple of another matrix in the basis, thus simplifying all the calculations. (Analogy with the conventional $SO(3)$ generators is a more transparent way to see the power of this basis.)
But even if you chose a less clever basis than the Pauli matrices, you could derive the same results. The equations could just become a bit more cumbersome.
Line crossing etc. is first observed at degeneracy 2. However, one may also study $3\times 3$ or $N\times N$ matrices. Conventional bases of the Hermitian matrices are called the Gell-Mann matrices in the case of $SU(3)$ - because Gell-Mann generalized the Pauli matrices when he studied the strong force where $SU(3)$ enters in two ways. The $N\times N$ Hermitian matrices depend on $N^2$ real parameters. One usually takes the identity matrix to be one of the basis vectors; the remaining $N^2-1$ vectors are "generators of $SU(N)$".