How to correct this free body diagram for two masses on eachother on a table? Assuming this arrangement,

And that the upper mass is lighter than the lower one, And that the coefficient of friction between the two masses is $\mu$, And the table is frictionless, this is the free body diagram I have drawn.

But it must have an error. Because if the force applied isn't more than maximum static friction, then the net force applied to M will be zero. But we know that the system will move.
How to correct it?
 A: I'll provide a complete solution to the case where the friction force is fully developed. I'm not entirely sure where your confusion arises, but I hope this will make the situation clear to you. Mechanics is difficult and there are lots of places you can make mistakes (I really hope that I haven't messed up myself).
First assume that all motion will be horizontal, so $\vec{a}_1 = a_1 \hat{i}$, $\vec{a}_2 = a_2 \hat{i}$. Then draw the free-body diagram and write down the force equations for the two blocks and the relation between friction and normal force:

Now that we have five equations, we can solve for our five unknowns ($a_1$, $a_2$, $N_1$, $N_2$ and $F_f$):
$$ \left.\begin{aligned}
    N_1 - mg &= 0 \\ F_f &= ma_1 \\
    N_2 - Mg - N_1 &= 0 \\ F - F_f &= Ma_2 \\ F_f &= \mu N_1
\end{aligned}\right\} \implies
\left\{\begin{aligned}
    N_1 &= mg \\ N_2 &= (M+m)g \\ F_f &= \mu mg \\ a_1 &= \mu g \\
    a_2 &= \frac{F - \mu mg}{M}
\end{aligned}\right. $$
In the case where the friction isn't fully developed we don't have $F_f = \mu N_1$ anymore, just $F_f < \mu N_1$. So just four equations but still five unknowns. But we can get another equation if we realise that $F_f$ is just the right value to prevent any relative motion between the blocks; $v_1 = v_2$ always and so $a_1 = a_2$. Now we have five equations again, and can solve the resulting system:
$$ \left.\begin{aligned}
    N_1 - mg &= 0 \\ F_f &= ma_1 \\
    N_2 - Mg - N_1 &= 0 \\ F - F_f &= Ma_2 \\ a_1 &= a_2
\end{aligned}\right\} \implies
\left\{\begin{aligned}
    N_1 &= mg \\ N_2 &= (M+m)g \\ F_f &= \frac{mF}{M+m} \\
    a_1 &= \frac{F}{M+m} \\
    a_2 &= \frac{F}{M+m}
\end{aligned}\right. $$
As a bonus, we can actually find the value of $F$ for which the blocks start gliding relative to each other by requiring that $F_f = \mu N_1$ in this solution (or alternatively, requiring that $a_1 = a_2$ in the previous solution):
$$ F_f = \mu N_1 \implies \frac{mF}{M+m} = \mu mg \implies
F = \mu(M+m)g $$
or alternatively
$$ a_1 = a_2 \implies \mu g = \frac{F - \mu mg}{M} \implies
F = \mu(M+m)g. $$
Both ways give the same result (we've just solved the system of all six equations with $F$ as a new unknown).
A: 
 Because if the force applied isn't more than maximum static friction, then the net force applied to M will be zero

This is incorrect. Nothing in this problem says that the static friction must be overcome. 
There is only static friction between the two boxes, and if it reached it's maximum, then the top box would start sliding and not holding on. It's not about the motion of the bottom box but rather about if the two are able to hold on to each other. 
So the short answer: Nothing is wrong in the free body diagram. The mistake is to assume static friction to equal the force $F$.
A: 1) Draw a free body force diagram for each mass.
2) Think about the frictional force between the two masses. Is it kinetic in nature or static in nature?
3) Remember that the only horizontal force acting on the upper mass is the frictional force regardless of nature.
For the bottom mass:
$$F-f=m_1a_1$$
For the top mass, $f=m_2a_2$ and $\mu_?m_2g=m_2a_2$.
Then:
$$F-\mu_?m_2g=m_1a_1$$
$$F-m_2a_2=m_1a_1$$
