Time derivative with respect to an observer moving with velocity $mathbf{v}$

Here $frac{df}{dt}$ is a total derivative. A total derivative is a derivative with respect to all of its variables. You probably already know this but what is often overlooked is that the total derivative is defined on a one-dimensional path. The function $f(x,y,z,t)$ is defined on $mathbb R^4$, which is 4-dimensional, so you have to write $x,y,z$ as a function of $t$ before you can take the total derivative. So you get $$frac{d}{dt}f(x(t),y(t),z(t),t)=frac{partial f}{partial x}frac{dx}{dt}+frac{partial f}{partial y}frac{dy}{dt}+frac{partial f}{partial z}frac{dz}{dt}+frac{partial f}{partial t}$$ So what have we gained by writing this so explicitly? Now we can notice the author has written $mathbf x$ as a function of $t$. He has defined some path through space which the observer travels along. So compare this to the total derivative for an observer at a fixed point $mathbf x(t)=mathbf x_0$: $$left.frac{df}{dt}right|_{mathbf x(t)=mathbf x_0}=frac{partial f}{partial t}(mathbf x_0)$$ This is probably closer to what you had in mind.

To answer your second question the formula stays the same for a time dependent velocity of the observer. To see this align the velocity of the observer with one of the axes. For example the x-axis. Now the formula is just the ordinary chain rule $frac{df}{dt}=frac{df}{dx}frac{dx}{dt}$. There's no acceleration in there.