Edit: To clarify the question, I am asking why we are justified in calling a continuous symmetry a symmetry of a system when it changes the Lagrangian by a total derivative of a function of $t, q(t)$ and $\dot{q}(t)$. As stated below, it is not clear to me that this would leave the equations of motion unchanged. A simple example from elementary physics is time translation $t \to t + \delta t$, whence $\delta L \propto dL/dt$ (in the infinitesimal case). The references to Noether's theorem are to give context to the question.
Edit 2: To emphasize (thanks to @Qmechanic), my question is about the common statement, for example found on p. 17 of Peskin and Schroeder but generic, that if $L$ changes by a total time derivative of some $F(t,q,\dot{q})$ under a symmetry, then that symmetry is a symmetry of the problem. I do not see how this follows when $F$ is a function of $\dot{q}$.
Almost everywhere I have seen it, Noether's theorem is said to be applicable when the Lagrangian changes by either a total time derivative in the discrete case, or by a four-divergence in the field case, of some function I will call $F$. My confusion holds in the discrete case, so I will use that since it is simpler. So we have, in most generality, if \begin{equation} L(t,q(t),\dot{q}(t)) \to L(t,q(t),\dot{q}(t)) + \frac{d F(t,q(t),\dot{q}(t))}{dt} \end{equation} under some continuous symmetry applied to $t$ and $q$, we say (although I suspect this may be sloppy) that the action and equations of motion are invariant under this continuous symmetry (and there is a conserved current, but that is outside this discussion).
It is clear why this implies the equations of motion remain unchanged under this symmetry if $F$ is not a function of $\dot{q}$ from the derivation of the Euler-Lagrange (EL) equations from the least action principle; since we fix the variation in $q$ to be zero at the endpoints, this function will always contribute nothing to the variation of the action.
I do not understand why this still holds if $F$ is a function of $\dot{q}$, since in the standard derivation of the EL equations we make no statement about the variation of $\dot{q}$ at the endpoints, and then a total derivative of a function of velocity can contribute to variations of the action: \begin{equation} \delta S = F(t,q(t),\dot{q}(t))|_{t_0}^{t_f} \ne 0, \end{equation} in general changing the equations of motion.
A good example of $F$ depending on $\dot{q}$ is the continuous symmetry of time translation, where $F \propto L$.
Possible solutions:
Do we have to introduce boundary conditions so that $\dot{q}$ is not varied at the endpoints as well? This seems strange and unphysical, as we will be imposing double the regular boundary conditions (4 vs 2 for a free particle, for example; normally we fix only the two endpoints).
Do we impose conditions on $F$ itself, so that $\delta S$ above is forced to be zero? This seems strange as well, especially given the example of $F \propto L$, as this in turn constrains $L$.
Added in edit: Does "symmetry of the system" not necessarily mean that the equations of motion are left unchanged?
Or does this mean that we do not necessarily treat this continuous symmetry as a symmetry of the equations of motion (so it is wrong to say that it in general leaves the action or equations of motion invariant), but only use Noether's theorem to find the conserved current (the current can be derived without assuming anything about the action)?
I suspect these are all slightly off-base, and thank you for your responses.
