A spherical wave is a solution of the three-dimensional wave equation of the form $u(r, t)$, where $r$ is the distance to the origin (the spherical coordinate). The wave equation takes the form $u_{t t}=c^{2}\left(u_{r r}+\frac{2}{r} u_{r}\right) \quad$ ("spherical wave equation"). (a) Change variables $v=r u$ to get the equation for $v: v_{t t}=c^{2} v_{r r} .$
Assume that $v(r,t)=ru(r,t).$
$v_t=ru_t$
$v_{tt}=ru_{tt}$
$v_{r}=u(r,t)+ru_r$
$v_{rr}=u_r+u_r+ru_{rr}=2u_r+ru_{rr}$
$\dfrac{v_{rr}}{r}=2\dfrac{u_r}{r}+u_{rr}$
$v_{r}=u(r,t)+ru_r$
$v_{rr}=u_r+ru_{rr}+u_r$
$v_{rr}=2u_r+ru_{rr}$
I see no way to conclude that $v_{tt}=c^2v_{rr}$ or $u_{tt}=c^2(u_{rr}+\dfrac{2}{r}u_r)$
The 3D wave equation is
(1) $ u_{tt}= \triangle u$ where the term on the right is the laplacian. If $u$ is a function only of the radial spherical coordinate then
$\triangle u= u_{rr} + \frac{2}{r} u_r $ or equivalently
(2) $ \triangle u= r^{-2} ( r^2 u_r)_r$
Now make the substitution $v= ru$ so that
(3.A) $ u=v/r$ and
3.B $u_{tt}= v_{tt}/r$.
Deduce from 3.A that
(4) $ u_r= \frac { v_r}{r} - r^{-2} v$
Thus (2) becomes
$ \triangle u=r^{-2} [ r v_r -v]_r $
where the bracketed term simplifies to $[ r v_r -v]_r = rv_{rr} $
Thus (2) becomes $ \triangle u= r^{-2} rv_{rr} = r^{-1} v_{rr} $
Now substitute this into (1) to see that the wave equation becomes (4) $ r^{-1}v_{tt}= r^{-1} ( v_{rr})$
and conclude that $v_{tt}= v_{rr}$
P.S. For simplicity I just assumed $c=1$ to make the typing easier.