Why does this one move faster?

Considering $m_1$ and $m_2$ as a system, there is a net horizontal force $F$ acting on them so their centre of mass must accelerate with acceleration

$displaystyle a_0 = frac F {m_1+m_2}$

This is true regardless of whether or not $m_1$ moves relative to $m_2$.

The frictional force which opposes $m_1$ moving relative to $m_2$ has a maximum value of $mu m_1 g$. If $F - mu m_1 g le m_1a_0$ then $m_1$ will not move relative to $m_2$ and both blocks will accelerate with acceleration $a_0$. Substituting the expression for $a_0$ above, we see that this condition becomes

$displaystyle F - mu m_1 g le F frac {m_1} {m_1+m_2} displaystyle Rightarrow F frac {m_2} {m_1+m_2} le mu m_1 g displaystyle Rightarrow F le mu g (m_1+m_2) frac {m_1}{m_2}$

In this case, the frictional force is

$displaystyle F frac {m_2} {m_1+m_2}$

and we can see that (by Newton's Third Law) this acts on $m_2$ to accelerate it with acceleration $a_0$ too. In effect, the force $F$ is divided between $m_1$ and $m_2$ in proportion to their masses, so that they both have the same acceleration.

On the other hand, if

$displaystyle F > mu g (m_1+m_2) frac {m_1}{m_2}$

then the net force on $m_1$ is $F - mu m_1 g$ so $m_1$ accelerate with acceleration

$displaystyle a_1 = frac F {m_1} - mu g$

and the force on $m_2$ is $mu m_1 g$ so $m_2$ accelerates with acceleratiion

$displaystyle a_2 = mu g frac {m_1}{m_2}$

and we have

$displaystyle a_1 > mu g frac {m_1+m_2}{m_2} - mu g displaystyle Rightarrow a_1 > mu g frac {m_1}{m_2} displaystyle Rightarrow a_1 > a_2$

Also note that the acceleration of the centre of mass of the system is the weighted sum of $a_1$ and $a_2$ which is

$displaystyle frac {m_1a_1+m_2a_2} {m_1+m_2} = frac {(F - mu m_1 g) + mu m_1 g} {m_1+m_2} = frac {F} {m_1+m_2} = a_0$

as we expect.