We are solving for the sound wave caused by the motion of a spherical piston. In a spherically symmetric case, the classical wave equation reads: $$\phi_{tt}=a_0^2\Big(\phi_{rr}+\frac{2\phi_r}{r}\Big),\tag{1}$$ where $\phi$ is the velocity potential. The general solution for the forward moving wave is given by: $$\phi=\frac{1}{r}f(r-a_0t),$$ where $a_0$ is the constant speed of sound. The radial velocity $u$ is therefore: $$u=\frac{\partial\phi}{\partial r}=\frac{1}{r^2}f(r-a_0t) - \frac{1}{r}f'(r-a_0t).$$ If the piston motion is given by $R=R(t)$, then using the boundary condition for velocity at the piston $u(R(t),t)=dR/dt$, we get from the previous equation: $$\dot{R}=\frac{1}{R^2}f(R-a_0t) - \frac{1}{R}f'(R-a_0t),\tag{2}$$ which is a first-order ordinary differential equation that we can solve for $f$.
Well so it goes. However, the one thing I find annoying about this is the last equation. Namely, the term $f'(R-a_0t)$ is not the derivative of $f$ w.r.t $(R-a_0t)$, but rather the derivative of $f$ w.r.t $(r-a_0t)$, evaluated at $r=R$. Thus, can we solve this as an ODE even though the differentiation is not made w.r.t the same independent variable of the equation?
Edit: Replying to the comment by Mattos, $\phi$ represents the general form of a wave transported in the positive $r$ direction (i.e radially outwards), and it can be easily shown that it satisfies the wave equation, namely: $$\phi_t=-\frac{a_0}{r}f'(r-a_0t),$$ $$\phi_{tt}=\frac{a_0^2}{r}f''(r-a_0t),$$ $$\phi_r=\frac{1}{r}f'(r-a_0t) - \frac{1}{r^2}f(r-a_0t),$$ $$\phi_{rr}= \frac{1}{r}f''(r-a_0t)-\frac{2}{r^2}f'(r-a_0t)+\frac{2}{r^3}f(r-a_0t),$$ and substituting these values into equation (1) yields $$\frac{a_0^2}{r}f''(r-a_0t)=\frac{a_0^2}{r}f''(r-a_0t),$$ thus satisfying the equation.
As for the other question, since $R(t)$ is given, equation (2) can be solved for $f$ up to a constant.