Short diagrammatic answer. Because states which are orthogonal as vectors, are represented as antipodal points on the Bloch sphere — orthogonal states of qubits are as far from each other as possible and represent a sort of 'opposite' of each other, and the Bloch sphere represents this idea.
Longer mathematical answer. Consider the vector-valued function $\def\ket#1{|#1\rangle}$ $$ \ket{\psi_{\theta,\phi}} = \cos(\theta/2) \ket0 + \mathrm e^{i \phi} \sin(\theta/2) \ket{1}. $$ This function has a period of $4\pi$ in the argument $\theta$, as a result of that division by 2; this might be part of what you are finding strange. However, when we consider quantum state vectors, we are not actually interested in the vector itself, as we are in the equivalence class of all vectors which are the same up to scalar factors (i.e. global phases).
Because $\ket{\psi_{\theta,\phi}} \ne \mathbf 0$, we always have $\ket{\psi_{\theta,\phi}} \ne \ket{\psi_{\theta+2\pi,\phi}} = -\ket{\psi_{\theta,\phi}}$; but we do have $\ket{\psi_{\theta,\phi}} \propto \ket{\psi_{\theta+2\pi,\phi}}$ instead. 976
That is, adding $2\pi$ to the angle $\theta$ gives you a vector $\ket{\psi_{\theta+2\pi,\phi}}$ which represents the same state. Because we only care about the state-vector inasmuch as it represents a state, this vector-valued function with peridod $4\pi$ is used to stand for a state-valued function with period $2\pi$.
This becomes clearer if we use density operators to represent states, because global phases of state vectors do not affect the density operator. Define the operator $$ \rho_{\theta,\phi} = |\psi_{\theta,\phi}\rangle!\langle\psi_{\theta,\phi}| = \begin{bmatrix} \cos^2(\theta/2) & \mathrm e^{-i\phi}!\cos(\theta/2)\sin(\theta/2) \ \mathrm e^{i\phi} !\cos(\theta/2)\sin(\theta/2) & \sin^2(\theta/2) \end{bmatrix};$$ using the double-angle formulae $$ \begin{gather} \cos(\theta) = 2\cos(\theta/2)^2 - 1= 1 - 2\sin(\theta/2)^2\ \sin(\theta) = 2\cos(\theta/2)\sin(\theta/2) \end{gather} $$ we may re-write $\rho_{\theta,\phi}$ as $$ \rho_{\theta,\phi} = \frac{1}{2}\begin{bmatrix} \cos(\theta) + 1 \;&\; \mathrm e^{-i\phi} \sin(\theta) \ \mathrm e^{i\phi} \sin(\theta) \;&\; 1 - \cos(\theta) \end{bmatrix} ,$$ which is not the best representation for thinking about qubits, but certainly makes it very clear that the density operator corresponding to $\ket{\psi_{\theta,\phi}}$ is periodic in $\theta$ with period $2\pi$.
As icing on the cake, note that $\rho_{\theta,\phi}$ is Hermitian by construction. So are the Pauli spin operators $\mathbf 1, X, Y, Z$, and those four are linearly independent. That means that we can decompose $\rho_{\theta,\phi}$ as a linear combination of those four operators. We may easily show, from the last expression for $\rho_{\theta,\phi}$, that $$ \rho_{\theta,\phi} = \tfrac{1}{2}\Bigl(\mathbf 1 + \cos(\phi)\sin(\theta) X + \sin(\theta)\sin(\phi) Y + \cos(\theta) Z \Bigr). 804