Imagine your kinetic energy $T$ and your potential energy $V$. The kinetic energy, $T$, depends on your velocity, the potential energy, $V$, depends on your location. So if you want to know your Lagrangian $L=T-V$ then you need to know both.
So imagine a plane where the $x$-axis tells your position and the $y$-axis tells you your speed, the Lagrangian $L=T-V$ is a function of the plane. And you can imagine looking at how the Lagrangian changes as a function if you move around this plane in the $x$ direction or in the $y$ direction. These are exactly what $\partial L / \partial x$ and $\partial L / \partial v$ are meant to denote. They do not denote what happens in time or as the particle moves. it is about how the Lagrangian would have been different had the position or the speed been different.
In three dimensions, you can imagine a 6d space, with the first three directions telling you the three parts of the position vector, and the next three giving the three parts of the velocity. The Lagrangian $L=T-V$ is a function of the 6d space. And you can imagine looking at how it changes if you had been at a different $x$ or at a different $\dot x$. The "rate" of change for these different position/velocity values are exactly what $\partial L / \partial x$ and $\partial L / \partial \dot x$ are meant to denote.
You must think of the Lagrangian $L=L(x,y,z,\dot x, \dot y, \dot z)$ as a function of hypothetical values of position $(x,y,z)$ and velocity $(\dot x, \dot y, \dot z)$, and take partials as a function, i.e. holding all the other variables (inputs to the function) constant. It is not motion in real space it is a hypothetical change in what the values of position or velocity could have been, you want to see how the Lagrangian would have been different for different initial positions and different initial velocities.
edit to respond to information in the comments
The Lagrangian could be different if the initial position were different. The Lagrangian could be different if the initial velocity were different.
But if you changed only the initial velocity, then the potential energy $V=V(x)$ doesn't change so $\partial V /\partial v =0$.
Similarly if you changed only the initial position, then the kinetic energy $T=T(v)$ doesn't change so $\partial T /\partial x =0$.
Therefore when you check to see how the Lagrangian $L=T-V$ changes when you adjust the initial position you get $\partial L/\partial x=0-\partial V/\partial x$ and when you check to see how the Lagrangian $L=T-V$ changes when you adjust the initial velocity you get $\partial L/\partial v=\partial T/\partial v-0$
I don't think it is always helpful to think about derivatives of $x$ with respect to $v$ (or vice versa), just think about how the potential energy changes if your initial velocity changes or think about how the kinetic energy changes if your initial position changes.