I think I understand your question. How is it possible, that even though the first ball deforms and stores elastic energy during the collision, it suddenly ends up with no motion after. The short answer is that some elastic energy is temporarily stored in the left-most ball during the collision, but somehow during the collision, the combination of forces, compression and relative motion makes it so all this stored energy is finally transmitted to kinetic energy in the right-most ball. The quickest way to argue why it must be so is by assuming conservation of energy and momentum, as is typically done in any introductory mechanics course. However, it seems that you would like to "see" the process unfold during the process during the collision.
Let us model the balls as elastic springs obeying <PERSON>'s law. Let $x_1,x_2$ be the positions, $v_1,v_2$ the velocities, $\Delta x_1, \Delta x_2$ the widths, and $m_1,m_2$ the masses of the left-most and right-most ball, respectively.
We assume the balls have equal size and spring constant so that the equilibrium length of both is $\Delta x_0$ and the spring constant is $k$. At the start of the collision, we set $x_1=0$ so that $x_2=\frac{\Delta x_0}{2} + \frac{\Delta x_0}{2}$ and the initial velocities are $v_1 = v_0$, $v_2=0$. During the collision the balls deform, store elastic energy, and exert a force on each other.
Due to N3, the force $F_{21}$ exerted by ball 2 on ball 1 is equal to $-F_{12}$, where $F_{12}$ is the force exerted by ball $1$ on ball $2$. Since the forces are related to the deformation of the balls, we have
$$ F_{12} = -k_1 (\Delta x_1 - \Delta x_0) $$
$$ F_{21} = k_2 (\Delta x_2 -\Delta x_0) $$
Equating $F_{12} = -F_{21}$, we find
$$ k (\Delta x_1 - \Delta x_0) = k (\Delta x_2 -\Delta x_0) $$ $$ \Delta x_1 = \Delta x_2 \equiv \Delta x $$
Additionally, we have
$$ x_2-x_1 = \frac{\Delta x_1}{2} + \frac{\Delta x_2}{2} = \Delta x. 512
$$
From N2, the equations of motion are
$$ a_1 = \frac{k}{m_1} (\Delta x_2 - \Delta x_0) = \frac{k}{m_1} (x_2-x_1 - \Delta x_0) $$ $$ a_2 = -\frac{k}{m_2} (\Delta x_2 - \Delta x_0) = -\frac{k}{m_2} (x_2-x_1 - \Delta x_0). $$
where $a_1, a_2$ are the accelerations of balls 1 and 2, respectively. I don't know if these equations can be solved analytically, but for the purposes of this post, I wrote a midpoint Euler script to solve them numerically. Below is the result from setting $k = 5 N/m$, $\Delta x_0=1 m$, $v_0=3 m/s$, $m_1 = m_2 = 1 kg$:
As we see from the third column (since $\Delta x_1 = \Delta x_2 = \Delta x = x_2-x_1$), both balls are compressed, and thus store elastic energy during the collision. The first ball does move during the collision and ends up at approximately $x_1\sim 0.6m$ after the collision. All the kinetic energy from the first ball is given onto the second, as seen in the middle figure. I made a simple animation of how it looks when two balls of equal mass collide.
The fact that the first ball comes to a complete stop is due to the masses being equal. If we increase the mass of the first ball to $m_1 = 2kg$ (keeping $m_2=1kg$), we get the following result
where, as you can see, the both balls continue along after the collision. For completeness, if we double the mass of the second ball and keeping the first at its original mass, i.e. 512