Invariance of Lagrange on addition of total time derivative of a function of coordiantes and time
Even if you change frames, the physics is still the same and the particle will follow the same path, no? And there is certainly more than one way to change the Lagrangian without affecting the path of least action--add any combination of total time derivatives to it.
When I said it would follow the same path, I meant the same path after you take into account the fact that you shifted frames. If $q_1$ and $q_2$ label the same point even after you shift frames (so that in new coordinates $q_1=q^{new}_1−ϵt$ and etc.) then the particle will be at $q_2$ at $t_2$ if it was at $q_1$ at $t_1$.
I mean that if the particle starts at time $t_1$ at $q_1$ it DOES end up at $q_2$ at time $t_2$ provided you take into account how the points look different because of the new frame. The path taken by the particle must be the same; physics doesn't depend on the inertial frame you are in and this is the point Landau is making. If you think that in frame K' the particle doesn't end up at $q_2$ at time $t_2$ then it is purely because the points are labeled differently in this frame. It also has nothing to do with v being infinitesimal; that doesn't matter.
As for the other question, you can also multiply the whole Lagrangian by a constant. That's kind of obvious though. Basically you need the kinetic energy term, and the only way you can further modify it is by adding terms right? If you multiplied the Lagrangian by a non-constant term, for instance, the form of the kinetic energy term would change. Then he rules out what terms you can add.
Maybe if you go to this page and look under the section "Is the Lagrangian unique?" it will help:
en.wikibooks.org/wiki/Classical_Mechanics/Lagrange_Theory
Basically you change the Lagrangian by adding terms, and it can be proved that ONLY if this term is a total time derivative of a function of coordinates and time that the action is still extremized. In other words, no, it is not possible to keep the path of least action by adding any other function.
Consider the principle of least action $\delta S = 0$, where the action $S$ is given by: $$S= \int_{t_1}^{t_2}L(\mathbf{q}(t),\dot{\mathbf{q}}(t),t)\ \text{d} t$$ from this we can derive Lagrange's Equations. Now let $G\in C^1(t_1,t_2)$, that is $G$ is an at least once differentiable function on $(t_1,t_2)$. Then $$S' = \int_{t_1}^{t_2}L(\mathbf{q}(t),\dot{\mathbf{q}}(t),t) +\dfrac{\text{d}G}{\text{d}t}\ \text{d} t=\int_{t_1}^{t_2}L(\mathbf{q}(t),\dot{\mathbf{q}}(t),t)\ \text{d} t + \Delta G = S + \text{const}$$ Thus $\delta S' = \delta S$ and $L(\mathbf{q}(t),\dot{\mathbf{q}}(t),t) +\dfrac{\text{d}G}{\text{d}t}$ is an equivalent Lagrangian of the system and will produce all the same dynamics.