Schrodinger equation just says that
$$
\hat{H}\left| \psi(t) \right\rangle = i\hbar \frac{\partial}{\partial t} \left| \Psi(t) \right\rangle
$$
It is an operator equation, in order to transform it into a PDE of a c-number function (the wave function) you have to project the Hilbert space vector $\left| \Psi(t) \right\rangle$ into its components. Now the hilbert space for a spineless particle in an infinite dimensional space, one whose bases is $\left| \vec{x} \right\rangle$. So we project to get
$$
\int d^3 x' \left\langle \vec{x} \right |\hat{H}\left| x'\right\rangle\left\langle x'\right.\left| \psi(t) \right\rangle = i\hbar \frac{\partial}{\partial t} \left\langle \vec{x} \right |\left. \Psi(t) \right\rangle
$$
which for local potentials becomes
$$
\hat{H}_x\Psi(\vec{x},t) = i\hbar \frac{\partial}{\partial t} \Psi(\vec{x},t)
$$
Where $\hat{H}_x\equiv\left\langle H \right\rangle_x$ Now if the particle has spin, or any new quantum number that provides the hilbert space with additional dimension "on top of" that of $\left| \vec{x} \right \rangle$, i.e. $\mathcal{H}_{\text{free}}\otimes\mathcal{H}_n$, you have to project onto that as well
$$
\sum_{n'} \int d^3 x' \left\langle \vec{x},n \right |\hat{H}\left| x',n'\right\rangle\left\langle x',n'\right.\left| \psi(t) \right\rangle = i\hbar \frac{\partial}{\partial t} \left\langle \vec{x},n \right |\left. \Psi(t) \right\rangle
$$
which again for local potentials becomes
$$
\sum_{n'} \hat{H}_{x,nn'} \Psi_{n'}(\vec{x},t) = i\hbar \frac{\partial}{\partial t} \Psi_n(\vec{x},t)
$$
In the case of spin 1/2 particle, $\mathcal{H}_n$ is two dimensional and therefor the above equation represents a vector equation, or two equations for each spin component. And $\hat{H}_{x,nn'}$ is just a $2\times2$ matrix, where it's usually the free hamiltonian multiplied by $\mathbb{1}_2$ plus some combination of the $2\times 2$ pauli matrices
Notice that $\Psi_n(\vec{x},t)$ is a two component vector (a spinor) with index $n$ in the same way as it is an infinite component vector with "index" $x$.
