Adding a total time derivative term to the Lagrangian

Well you just showed ${d over dt } { partial L' over partial dot q}- { partial L' over partial q}= {d over dt } { partial L over partial dot q}- { partial L over partial q}=0$ right? ${d over dt } { partial L over partial dot q}- { partial L over partial q}=0$ is the equation of motion for $q$, in other words this equation means exactly that: $L$ and $L'$ give the same equation of motion for q.

Here is another way to think about it, using the variational principle version of the Euler-Lagrange equations.

The action of $L$ and $L'$ differ by $dot F$.

$$mathcal S = int^{t_f}_{t_i} dt L$$

$$mathcal S' = int^{t_f}_{t_i} dt L' = int^{t_f}_{t_i} dt (L+dot F) = mathcal S + F(t_f)-F(t_i) $$

Since $F(t_f)-F(t_i)$ is a constant, the paths that extremize $mathcal S$ and $mathcal S'$ are the same.

If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says:

$frac{partial dot{F}}{partial dot{q}}=frac{partial F}{partial q}$.

This equation says, although not obviously, that the last two terms in the fourth equation in the question, notably this equation:

$ frac{d}{d t} frac{partial L}{partial dot{q}}-frac{partial L}{partial q}+frac{d}{d t} frac{partial}{partial dot{q}} frac{d F}{d t}-frac{partial}{partial q} frac{d F}{d t}=0 $.

are in fact equal and thus cancel. So you are left with

$frac{d}{d t} frac{partial L}{partial dot{q}}-frac{partial L}{partial q} = 0$

From that you can get the equations of motion, just as you would with $L'$. So $L'$ and $L$ give the same equations of motion.

But to get why the time derivative of $F$ is important, and not just $F$, lets start with the third term which is $frac{d}{d t} frac{partial}{partial dot{q}} frac{d F}{d t}$ and write it as $frac{d}{d t} frac{partial}{partial frac{d}{dt}{q}} frac{frac{d}{dt} F}{1}$. Now you can see that we are taking the partial of the rate of change of $F$ with respect to the rate of change of $q$. Also not stated in the question, it is required that $F$ is a function of $t$ and $q$. That is $F=F(q,t)$

So we can dump the rate of change part and just keep the derivative of $F$ w.r.t. $q$ and get $frac{d}{d t} frac{partial}{partial {q}} F$, which is $ frac{partial}{partial {q}} frac{d F}{d t}$ which is the same as the fourth term and therefore they cancel one another.

Without the $d/dt$ in front of $F$ this would not have worked. So, adding to the Lagrangian a total time derivative of a function of $q$ and $t$ does not change the equations of motion

This can be easily calculated as seen in the attached image.