If one takes the step size of an $n$-dimensional symmetric random walk to be infinitesimal, then the transition probability becomes the heat kernel. Thus, symmetric random walks are discrete, or microscopic, models of heat/diffusion.
The heat equation and wave equation are merely different in the time derivative. So what is the minimal discrete/microscopic model for wave propagations, analogous to random walks?
I'm not exactly sure what you're looking for, but here's how I think about this at a discrete level (this follows the Wikipedia article on the wave equation).
Consider a line of springs each of mass $m$ and length $h$, with spring constant $k$. The distance a spring, located at $x$, is displaced from equilibrium is denoted by $y(x)$.
The force of the spring at location $x+h$ is $$F = m \frac{d^2 y(x+h)}{dt^2}.$$
From Hooke's law, the mass balance on this spring is given by $$F = F^{x+2h}-F^{x}$$ where the superscript means the force exerted by all of the springs on that side of the spring under consideration.
Next, $$F = F^{x+2h}-F^{x}=k([y(x+2h)-y(x+h)]-[y(x+h)-y(x)]).$$ Finally, we take the number of springs to be $N$, with the total mass being $M=Nm$, the total spring constant being $K=k/N$ and the total length is defined as $L=Nh$.
Therefore, we have $$\frac{d^2 y(x+h)}{dt^2}=\frac{KL^2}{M}\frac{ y(x+2h)-2y(x+h)-y(x)}{h^2}.$$
Taking the limits $h \to 0, N\to \infty$ and defining $c^2 =KL^2/M$, we have the wave equation $$y_{tt}-c^2 y_{xx}=0.$$
The second order forward finite difference stencil is given by $$ f''(t) = \frac{f(t+2\Delta t)-2f(t+\Delta t) + f(t)}{\Delta t^{2}} $$ Define the update rules as $$ P(x,t + \Delta t) = pP(x-\Delta x,t) + qP(x+\Delta x,t) $$ $$ P(x,t+2\Delta t) = p\left[pP(x-2\Delta x,t) + qP(x,t)\right] + q\left[pP(x,t) + qP(x+2\Delta x,t) \right] $$ Using the finite difference stencil, Taylor expanding $P(x \pm \Delta x,t)$ and collecting terms we find \begin{align*} P(x,t+2\Delta t) - 2P(x,t + \Delta t) + P(x,t) = & \left(p+q\right)\left(p +q -1\right)P(x,t) + \\ & 2\left(-p^2 + q^2 + p -q \right)\Delta x \partial_{x}P(x,t) + \\ & \left(2p^2 + 2q^2 - p -q\right)\Delta x^{2}\partial_{xx}P(x,t) \end{align*} Using $ p+q=1$ and dividing with $ \Delta t^{2} $ we find the equation $$ \partial_{tt}P(x,t) = 4\left(p - \frac{1}{2}\right)^{2} \frac{\Delta x^{2}}{\Delta t^{2}}\partial_{xx} P(x,t) $$ So the same update rule which gives diffusion( see for example Diffusion coefficient for asymmetric (biased) random walk) gives the wave equation. Apparently it is all a matter of how to take the limit of $ \Delta x \rightarrow 0 $ and $ \Delta t \rightarrow 0 $.
