Adding a total time derivative term to the Lagrangian 
This is proof that $L'$ represents same equation of motion with $L$ through Lagrange eq.
I understand $L'$ satisfies Lagrange eq, but how does this proof mean $L'$ and $L$ describe same motion of particle?
In other words, why does total time derivative term which is added to $L$ make no difference in equation of motion?
 A: 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.
A: 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
A: You have seen that the substitution 
$$L\longrightarrow L':= L+\frac{\mathrm{d}F}{\mathrm{d}t}$$
does not change the Euler-Lagrange equations. Now, this happens because the time derivative satisfies the Euler-Lagrange equations identically. 
Let us consider a concrete example. Take the Lagrangian of a simple harmonic oscillator:
$$L_\text{HO}=\frac{1}{2}m\dot q^2-\frac{1}{2}m\omega^2q^2$$
which gives the Euler-Lagrange equations
$$\ddot q=-\omega^2 q$$
Now consider the modified Lagrangian
$$L'=\frac{1}{2}m\dot q^2-\frac{1}{2}m\omega^2q^2+\dot q=L_\text{HO}+\dot q$$
The Euler-Lagrange equations are obviously linear. Thus
$$\text{EL}[L']=\text{EL}[L_\text{HO}]+\text{EL}[\dot q]$$
As was shown above, $\dot q$'s Euler-Lagrange equation will vanish, but we can verify this:
$$\text{EL}[\dot q]:=\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial \dot q}{\partial\dot q}-\frac{\partial\dot q}{\partial q}=\frac{\mathrm{d}1}{\mathrm{d}t}-0=0$$
Thus, 
$$\text{EL}[L']=\text{EL}[L_\text{HO}]$$
i.e. the modified Lagrangian still implies $\ddot q=-\omega^2q$.
A: 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. 
A: This can be easily calculated as seen in the attached image.

