$\newcommand{\R}{\mathbb{R}}$HYPOTHESIS
Let $\Omega $ be an open, connected subset of $\R^n$, denote by $C^\infty_0(\Omega)$ the functions with compact support in $\Omega$.
Let $x_0 \in \mathrm{int} (\Omega)$ and suppose that $g\in L^2(\Omega)$ satisfies
$$(*)\quad \quad \int_\Omega \left(\Delta f(x)\right) \overline{g(x)} \ dx \ +\ f(x_0) = 0 $$
for all $f\in C_0^\infty(\Omega)$ and $$(**) \quad \quad g|_U \equiv 0 $$ where $U\subset \Omega$ is an open ball.
DEDUCTION
(*) implies that for all $f\in C_0^\infty(\Omega \setminus \{x_0\})$, $\int_\Omega \left(\Delta f(x)\right) \overline{g(x)} \ dx = 0$, which tells us that $\Delta g = 0$ weakly on $\Omega\setminus\{x_0\}$, and therefore by elliptic regularity ($g\in L^2(\Omega))$ , $g|_{\Omega\setminus \{x_0\}}$ is a classical solution, hence is in $C^\infty(\Omega\setminus\{x_0\})$.
Now (**) implies by unique continuation that $g|_{\Omega\setminus \{x_0\}}\equiv 0$ and thus $g|_{\Omega}\equiv 0$. But this is in contraddiction with $(*)$.
Is there some hidden mistake in my argument or really $(*)$ and $(**)$ are incompatible conditions?
I abstracted this from my original problem where instead of the operator $\Delta$ we have a Dirac operator and instead of $\Omega$ we have a closed manifold.
You are correct that these are incompatible conditions. The first hypothesis is the statement that $\Delta g = \delta_{x_0}$ in the sense of distributions. Therefore $g$ must be a fundamental solution of the Laplacian on the domain $\Omega$, which does not vanish on any open set
(in fact, it must be positive everywhere).I am wrong about the fundamental solution being positive -- this is only true on some manifolds (for example it is positive on $\mathbb{R}^3$ but not on $\mathbb{R}^2$). However the unique continuation principle you use is valid and, as $g$ is harmonic outside of $x_0$, you are correct that it cannot vanish on an open set.