TL;DR The term $k_3 L$ from your question is not part of the differential equations, but rather $k_3 \Delta L$ where $\Delta L$ is difference between the total length $L$ and sum of lengths of all three springs in relaxed state. Problems like this usually assume $\Delta L = 0$ although it is not explicitly stated by the problem definition. Below I show why you should not neglect the $\Delta L$ term.
If $x$ denotes position measured from the left wall (as you have indicated in your question), then your (differential) equations for motion are not correct. Here is a simple check: for $x_1 = x_2$ the spring 2 is completely compressed, and yet your equations predict there is no force by the spring 2.
Let $x$ denote distance measured from the left wall. Then the left wall is at $x_L = 0$ and the right wall is at $x_R = L$. Nominal spring length (when there is no elongation) is denoted by $L_1$, $L_2$, and $L_3$. The spring forces are then defined as
$$F_1 = k_1 (L_1 - x_1) \qquad F_2 = k_2 (L_2 - (x_2-x_1)) \qquad F_3 = k_3 ((L-x_2)-L_3)$$
The equations of motion for both masses are
$$m_1 \ddot{x}_1 = F_1 - F_2 \qquad \text{and} \qquad m_2 \ddot{x}_2 = F_2 + F_3$$
which is expanded to
$$m_1 \ddot{x}_1 = k_1 (L_1 - x_1) - k_2 (L_2 - (x_2-x_1))$$
$$m_2 \ddot{x}_2 = k_2 (L_2 - (x_2-x_1)) + k_3 ((L-x_2)-L_3)$$
For reason that will be obvious later, let's now redefine the positions $x_1$ and $x_2$ as
$$\tilde{x}_1 = x_1 - L_1 \qquad \text{and} \qquad \tilde{x}_2 = x_2 - (L_1 + L_2)$$
Note that these substitutions do not affect the corresponding time-derivatives (accelerations) since spring length at relaxed state are constant
$$\frac{d^2}{dt^2} \tilde{x}_1(t) = \frac{d^2}{dt^2} x_1(t) \qquad \text{and} \qquad \frac{d^2}{dt^2} \tilde{x}_2(t) = \frac{d^2}{dt^2} x_2(t)$$
With these substitutions, the equations of motion now become
$$\boxed{m_1 \ddot{x}_1 = -k_1 \tilde{x}_1 + k_2 (\tilde{x}_2 - \tilde{x}_1)} \qquad \text{and} \qquad \boxed{m_2 \ddot{x}_2 = -k_2 (\tilde{x}_2 - \tilde{x}_1) - k_3 \tilde{x}_2 + k_3 \Delta L}$$
where $\Delta L = L - (L_1 + L_2 + L_3)$. The problem definition does not say anything about the spring lengths, so the affine term $k_3 \Delta L$ must be part of the equation. Problems like this are usually simplified such that sum of individual lengths equals total length in which case $\Delta L = 0$.
Although this is not part of the original question, for completeness of the solution I discuss it here. The system is in equilibrium when forces exerted by all springs are balanced, in which case $\ddot{x} = 0$
$$
\left\{
\begin{array}{ll}
(k_1 + k_2) \tilde{x}_1 - k_2 \tilde{x}_2 = 0 \\
k_2 \tilde{x}_1 - (k_2 + k_3) \tilde{x}_2 = -k_3 \Delta L
\end{array}
\right.
$$
from which solution for equilibrium immediately follows:
$$\tilde{x}_1 = \frac{k_2 k_3}{k_1 k_2 + k_2 k_3 + k_3 k_1} \Delta L \quad \text{and} \quad \tilde{x}_2 = \frac{k_2 k_3 + k_3 k_1}{k_1 k_2 + k_2 k_3 + k_3 k_1} \Delta L$$
When all three springs can be relaxed at the same time then $\Delta L = 0$ and the equilibrium is $\tilde{x}_1 = 0$ and $\tilde{x}_2 = 0$. In this case $\tilde{x}$ is a position relative to the equilibrium point.
Best Answer
Conceptually, the reason they are attached to walls is to remove rigid body motion from consideration. When they are tethered to the wall, there is no translation or rotation about the center of mass of the system.
So, when you break the connections to the walls and they are now free to drift through space, the number of admissible solutions increases.
If you connect one mass to the wall and leave the other end unconnected, then you remove the rigid body modes because it cannot rotate or translate and you're back to a similar, but different, problem as when both are connected to walls.
The equations of motion are easily derived for 3 bodies in both conditions when one draws the free body diagram (if deriving from a Newtonian approach) or through the application of Lagrange's equation or Hamilton's principle, depending on your choice of weapon.
But it's not that it's harder or impossible to have the three masses floating in space, it just adds more admissible modes to the solution which may or may not obscure the intent of the example/problem.