Spin-Angular Momentum Coupling: What component do $\hat{H}, \hat{x}$ and $\hat{p}$ act on? Consider a QM system with Hamiltonian $\hat{H}$, position operator $\hat{x}$, momentum operator $\hat{p}$ and angular momentum operator $\hat{L}$, all acting on a Hilbert space $\mathcal{H}$. Assume now that the system has spin, for example if we consider a hydrogen atom, and introduce the spin operator $\hat{S}$.
If we want to take into account the interaction of the two, we study the Hilbert space $\mathcal{H}\otimes\mathcal{H}$, as far as I understand. Here angular momentum acts on the first component and spin acts on the second.
Question: What component do $\hat{H}, \hat{x}$ and $\hat{p}$ act on?
 A: Consider first a spinless particle and denote, as you did, by $\cal H$ its Hilbert space. The operators $\hat x$ (position), $\hat p$ (momentum), $\hat H$ (hamiltonian) act on states of $\cal H$. The wavefunction of the particle in the state $|\psi\rangle$ is
$$\psi(\vec r)=\langle \vec r|\psi\rangle$$
Consider now a spin $1/2$ without any other degree of freedom. Its Hilbert space, say ${\cal H}_{1/2}$ is spanned by the two states $\{|\uparrow\rangle,|\downarrow\rangle\}$. The operators $\hat S_i$ ($i=x,y,z$) act on ${\cal H}_{1/2}$. In the basis $\{|\uparrow\rangle,|\downarrow\rangle\}$, it can be written as a $2\times 2$ matrix. If the spin is in the state $|\phi\rangle=a|\uparrow\rangle+b|\downarrow\rangle$, the probability amplitude of finding it in the $\uparrow$ state is
$$\langle\uparrow|\phi\rangle=a$$
Now, for a spin $1/2$-particle, i.e. having both translation and spin degrees of freedom, the quantum state belongs to the full Hilbert state ${\cal H}\otimes{\cal H}_{1/2}$. The probability amplitude of finding the particle at point $\vec r$ with a spin $\uparrow$ is
$$(\langle\vec r|\otimes\langle\uparrow|)(|\psi\rangle\otimes|\phi\rangle)
=\langle\vec r|\psi\rangle\ \!\langle\uparrow|\phi\rangle$$
The position operator $\hat x$ acts only on ${\cal H}$ (and should now be written $\hat x\otimes\mathbb I$):
$$\eqalign{
\hat x(|\psi\rangle\otimes|\phi\rangle)
&=(\hat x\otimes\mathbb I)|\psi\rangle\otimes|\phi\rangle\cr
&=(\hat x|\psi\rangle)\otimes(\mathbb I|\phi\rangle)\cr
&=(\hat x|\psi\rangle)\otimes|\phi\rangle\cr
}$$
The spin operators $\hat S_i$ act only on ${\cal H}_{1/2}$:
$$\eqalign{
\hat S_i(|\psi\rangle\otimes|\phi\rangle)
&=(\mathbb I\otimes\hat S_i)|\psi\rangle\otimes|\phi\rangle\cr
&=|\psi\rangle\otimes\hat S_i|\phi\rangle\cr
}$$
Some operators may act on both Hilbert state simultanously. It is the case of the spin-orbit interaction $\hat W\sim \vec L.\vec S$ that couples the spin and the angular momentum of the electron in an atom for example:
$$\eqalign{
   \vec L.\vec S|\psi\rangle\otimes|\phi\rangle
   &=(\vec L\otimes\mathbb I).(\mathbb I\otimes\vec S)|\psi\rangle\otimes|\phi\rangle\cr
   &=(\vec L|\psi\rangle)\otimes(\vec S|\phi\rangle)\cr
}$$
Note that such interaction causes the entanglement of translation and spin degrees of freedom. The eigenstates of $H$ are no longer a tensor product $|\psi\rangle\otimes|\phi\rangle$ but a linear superposition of such products.
PS: see also the answer How is the product $L\cdot S$ between orbital and spin angular momentum operators defined? Do they act on the same or different Hilbert spaces?
