Let $\mathcal{H}$ be a complex Hilbert space, $A$ a self-adjoint unbounded operator on $\mathcal{H}$ with domain $\mathcal{D}(A)$, and $g:\mathcal{D}(A)\rightarrow\mathbb{C}$ be an unbounded linear functional on $\mathcal{H}$ which is continuous with respect to the graph norm associated with $A$. I will denote its action on $\varphi\in\mathcal{D}(A)$ by $\langle g,\varphi\rangle$. The operator $A_0$ defined by \begin{equation} A_0=A\restriction\ker g \end{equation} is, as long as $g$ is unbounded, a densely defined symmetric operator on $\mathcal{H}$. Furthermore, its deficiency (or defect) indices, defined as usual by \begin{equation} d_\pm=\dim\mathrm{ran}(A_0\pm i)^\perp=\dim\ker(A_0^*\mp i) \end{equation} should be both equal to $1$. This is claimed in the classic reference Singular perturbations of differential operators by S. Albeverio and P. Kurasov, and it is the starting point to define rigorously rank-one singular perturbations of $A$: that is, operators corresponding to the formal expression "$A+\alpha\langle g,\cdot\rangle g$".
Now, unless I am losing something, in the book the authors do not prove explicitly that the deficiency indices of $A_0$ are in fact equal to $d_\pm=1$. They do show that the vectors $(A\pm i)^{-1}g$, to be intended as the vectors uniquely associated with the (bounded) forms \begin{equation} \varphi\in\mathcal{H}\mapsto\left\langle g,(A\pm i)^{-1}\varphi\right\rangle \end{equation} are indeed deficiency elements for $A_0$. This is easy: given $\varphi\in\mathrm{ran}(A_0\pm i)$, that is, $\varphi=(A\pm i)\psi$ for some $\psi\in\mathcal{D}(A_0)$ (i.e. $\psi\in\mathcal{D}(A)$ and $\langle g,\psi\rangle=0$), then \begin{equation} \left\langle g,(A\pm i)^{-1}\varphi\right\rangle=\left\langle g,\psi\right\rangle=0. \end{equation} However, this merely shows that $(A\pm i)^{-1}g$ are nonzero elements of the deficiency subspaces (whence $d_\pm\geq1$); one should show that any other element of these spaces is proportional to them, and the referenced book does not seem to provide a proof for that, merely stating that "the deficiency element is unique (up to multiplication by complex numbers)".
I am aware this is likely a very easy thing to prove. A possibility I was considering is the following: any element $\theta_{\pm i}\in\ker(A_0^*\mp i)$ must agree with $(A\pm i)^{-1}g$, as a bounded functional on $\mathcal{H}$, on the dense subspace $\mathrm{ker}\,g\subset\mathcal{H}$ (since both functionals must be identically zero on it), and this may somehow imply the desired relation. However, this does not seem to be enough.
The proof should be the following one, along the lines of Lemma II.2 in M. A. Astaburuaga et. al, J. Math. Phys. 63, 023502 (2022).
Since $g$ is continuous with respect to the graph norm on $\mathcal{D}(A)$, which is equivalent (see Section 1.2.2 of Albeverio and Kurasov's book referenced in the question) to the norms \begin{equation} \phi\mapsto\left\|(A\pm i)\phi\right\|, \end{equation} the functional $(A\pm i)^{-1}g$, defined by $\psi\in\mathcal{H}\mapsto\left\langle (A\pm i)^{-1}g,\psi\right\rangle:=\left\langle g,(A\mp i)^{-1}\psi\right\rangle$, is continuous on $\mathcal{H}$. The Riesz representation theorem implies that there exist two vectors $h_{\pm i}\in\mathcal{H}$ such that $h_{\pm i}=(A\mp i)^{-1}g$.
Now, let $\psi\in\mathcal{H}$. One has \begin{eqnarray} \langle h_{\pm i},\psi\rangle =0 &\iff& \left\langle g,(A\mp i)^{-1}\psi\right\rangle=0\\ &\iff&(A\mp i)^{-1}\psi\in\ker \langle g,\cdot\rangle\\ &\iff&\exists\phi\in\mathrm{dom}\,A,\,\langle g,\phi\rangle=0,\text{ s.t. } \psi=(A\mp i)\phi\\ &\iff&\exists\phi\in\mathrm{dom}\,A_0\text{ s.t. } \psi=(A\mp i)\phi\\ &\iff&\psi\in\mathrm{ran}(A_0\mp i). \end{eqnarray} Therefore, $\{h_{\pm i}\}^\perp = \mathrm{ran}(A_0\mp i)$ and thus, taking orthogonals, $\overline{\{h_{\pm i}\}}=\mathrm{span}\{h_{\pm i}\}=\mathrm{ran}(A_0\mp i)^\perp$. The linear span of a vector is one-dimensional, thus concluding the proof.