Consider a real-valued function $V=V\left ( x,y \right )$. Suppose $\left ( 0,0 \right )$ is a fixed point of the function f above.
In making an analogy to a physical system, say a spring without damping coefficient or a classical particle with energy E < V in a 'valley' of potential energy V.
In this case, the fixed point $x^{*}$ is an attracting fixed point and is also a minimum point. In the neighbourhood of this fixed point $x^{*}$, the function V of x in the neighbourhood of $x^{*}$ is such that $V\left ( x \right )>V\left ( x^{*} \right )$.
One may go further and see that for the real-valued function V, the set of level curve is the set $L_{c}$ of all points x of n-tuples such that $V\left ( x \right )=C$ where C is a constant. A physical analogy to this is the energy level of a particle at some height h in the valley. The energy level E of that particle at some height h is constant at any position on the plane.
Recalling that the gradient $\vec{\bigtriangledown }$ of a continuously differentiable function V is the greatest rate of change of the function V. Then, for the unit vector $\vec{u}$, $\vec{\bigtriangledown }.\vec{u}$ gives the greatest rate of change of the function V in the direction of the unit vector.
But what about the geometric intepretation of Lyapunov stability theorem which defines
$\dot{V\left ( x \right )}=\vec{\bigtriangledown }V\left ( x \right ).f\left ( x \right )$?
The interpretation is the following. When you have a function $V(\mathbf{x})$ and point $\mathbf{x}_0$, you might ask a question: how function value changes when you track it along some curve? So, you have a curve $\mathbf{x}(s)$ which passes through a point $\mathbf{x}_0$ (i.e., $\mathbf{x}(0) = \mathbf{x}_0$) and from calculus you know that in the first order the answer really depends only on a tangent vector $\mathbf{x}'(0)$: $$V(\mathbf{x}(s)) = V(\mathbf{x}_0)+ s \cdot \left (\nabla V(\mathbf{x}_0 ), \mathbf{x}'(0) \right ) + o(s).$$
The expression $\left (\nabla V(\mathbf{x}_0 ), \mathbf{x}'(0) \right )$ gives you the rate of change of function $V(\mathbf{x})$ when you are moving in direction $\mathbf{x}'(0)$ from point $\mathbf{x}_0$.
That's always true for any scalar function and any passing curve. But when you are dealing with system of ODEs, you can't move freely in any direction: system of ODEs defines a vector field $\dot{\mathbf{x}} = v(\mathbf{x})$ and it determines tangent vectors for all trajectories of the system.
So when we are discussing $$ \left ( \nabla V (\mathbf{x}_0), v(\mathbf{x}_0) \right )$$ this expression encodes the rate of change of this scalar function at point $\mathbf{x}_0$ when we are moving along trajectory.