In definition:
If the Jacobian of the dynamical system at an equilibrium happens to be a stability matrix (i.e., if the real part of each eigenvalue is strictly negative), then the equilibrium is asymptotically stable.
Here I have something unclear:
1- Does that mean the Jacobian of the open loop system, i.e., for the linearized model?
2- If yes, In case the linearized open-loop system was unstable and I have designed stabilizer for closed-loop of this system (linearized), Is the closed loop for the original nonlinear system with the stabilizer called asymptotically stable?
You perform linearization of a system, and then you decide about the stability of this system. If you linearize an open-loop system, then you decide about the stability of the open-loop system. If you linearize a closed-loop system, then you decide about the stability of the closed-loop system. Note that the open-loop system linearization and the closed-loops system linearization are (in general case) different.
UPD. Concerning the updated question. Let your open-loop system be $\dot{x}=f(x,u)$ , where $f(0,0)=0$. When you linearize your system at the origin, you have $$\dot{x}=Ax + Bu,$$ where $A=\frac{\partial f}{\partial x}(0,0)$ and $B=\frac{\partial f}{\partial u}(0,0)$. Suppose now that you propose $u=-Kx$, where $K$ is such that $A-BK$ is Hurwitz and the closed-loop linearization is stable. Then the nonlinear system is now $\dot{x}=f(x,-Kx)$, and its linearization at $x=0$ is now $\dot{x}=(A-BK)x$. Since the linearization is asymptotically stable, the closed-loop nonlinear system is locally asymptotically stable as well.