Choose the Hilbert space as $\mathcal{H}:=L^2(\mathbb{R})$ and define on it (or on a closed domain of it) the self-adjoint unbounded Schroedinger operator $H:=-\partial^2+V$ where $\partial$ is just the derivative operator and $V$ is the diagonal multiplication operator corresponding to the function $v:\mathbb{R}\to\mathbb{R}$ (which should probably satisfy some conditions to be stipulated later).
Since $\mathbb{R}=\left(-\infty,0\right]\sqcup\left(0,\infty\right)$, we have the orthogonal decomposition $\mathcal{H}=\mathcal{H}_L\oplus\mathcal{H}_R$ with $\mathcal{H}_L:=L^2(\left(-\infty,0\right])$ and $\mathcal{H}_R:=L^2(\left(0,\infty\right))$. Thus we should be able to write any operator $A$ as a two-by-two block form $A=\begin{bmatrix}A_{LL} & A_{LR} \\ A_{RL} & A_{RR} \end{bmatrix}$.
My question is, how does this proceed when choosing $A=H$? It is clear that $V_{LR}=V_{RL}=0$ since it is diagonal in the position basis, and $V_{LL},V_{RR}$ merely correspond to the restrictions of the function $v$. However, what about $-\partial^2$? How does it decompose? It must be that $(-\partial^2)_{LR}\neq0$ since otherwise the two sides of space would be uncoupled, which is not the problem we started with.
It seems like one could use some of the logic in solving the problem of a delta-well potential on the line, where one uses that the wave-function and its derivative are continuous. But here there is no delta function at $0$.
A simpler situation would be to define a Hamiltonian $H=-\frac{d}{dx^2}$ on smooth functions with support in $(0,1)$. This is a densely defined symmetric operator, but not "selfadjoint", which has a special meaning for unbounded operators. This can be understood physically by noting that $H$ can be extended to functions with support in $[0,1]$ with periodic boundary conditions or with "hard wall" boundary conditions. Each gives a different physical system with different eigenvalues and eigenvectors.
Your Hamiltonian will also not be "selfadjoint" and will have different extensions depending on choice of boundary conditions at $0$. For example, you can specify a particular extension by adding to the domain a smooth function which is non-zero at $0$. This is not in the domain of $A_{LL}$ or $A_{RR}$, so the block form is not well defined.