This may not be a direct answer to your question but it may give you an insight about Laplacian operator.
Now assume that we have a field $A(x,y,z)$. Let us expand this, for simplicity, around the origin. $$A(x,y,z)=A(0)+(x\partial_x+y\partial_y+z\partial_z)A(0)+(xy\partial_x\partial_y+xz\partial_x\partial_z+yz\partial_y\partial_z)A(0)+\frac{1}{2}(x^2\partial_x^2+y^2\partial_y^2+z^2\partial^2_z)A(0)+\text{Higher Order Terms}$$
where $A(0):=A(0,0,0)$. If we now take a volume integral over a cube around the origin, say $-\epsilon<x,y,z<\epsilon$, we get
$$\int A(x,y,z)dV=A(0)V+0+0+ \frac{4}{3}\epsilon^5(\partial_x^2+\partial_y^2+\partial_z^2)A(0)+\text{Higher Order Terms}$$ where $V:=(2\epsilon)^3$. We can define the average value of $A$ around the origin as
$$A_{ave}(0):=\frac{1}{V}\int A(x,y,z) dV$$hence our equation becomes
$$\Delta A(0)=\frac{6}{\epsilon^2}\left(A_{ave}(0)-A(0)-\text{Higher Order Terms} \right)$$
We can kill the Higher Order Terms by taking $\epsilon\rightarrow 0$, hence our equation becomes
$$\Delta A(0)=\lim_{\epsilon\rightarrow0}\frac{6}{\epsilon^2}\left(A_{ave}(0)-A(0)\right)$$
Therefore, Laplacian of a function at one point gives the difference between the functions value at that point and the average of the functions values in the infinitesimal neighborhood. Above, we used Cartesian coordinates for simplicity, hence the ugly $\frac{6}{\epsilon^2}$ in the front, but the result should be much more elegant in the spherical coordinates!
Back to your question: Since the difference of the average of surrounding and the point itself is actually related to the curvature (every point is average of its surroundings in a flat space), the RHS of the wave equation is indeed curvature induced force, and wave equation simply relates the change of the field due to the curvature (or more simply, it relates how a field changes in time as its values are not properly distributed in space, meaning that values at points being not equal to averages of their surroundings) .