I) For a single$^1$ spherically symmetric thin-shell collapse with (rest) mass $m$, the geometry inside the shell is flat Minkowski space, cf. e.g. this Phys.SE post. Outside the shell the geometry is Schwarzschild geometry with Schwarzschild radius $R_s=2E$ in natural units. Here
$$E~=~ m\sqrt{1+\left(\frac{dR}{d\tau}\right)^2} - \frac{m^2}{2R} \tag{3.70} $$
is the total energy of the thin shell, including the gravitational binding-energy. And here $\tau$ denotes the proper time of the shell. Note that the total energy $E$ is the gravitational mass of the shell, and a constant of motion.
II) If the formula (3.70) should be valid at $R=\infty$, we must demand that the total energy $E>m$ is bigger than the (rest) mass. In that case the Schwarzschild radius $R_s=2E$ cannot be $2m$, cf. OP's question (v2).
III) To reflect the collapse, the sign of the velocity is chosen negative
$$ \frac{dR}{d\tau}~=~-\sqrt{\left(e+\frac{m}{2R}\right)^2-1}~<~0, \qquad e~:=~\frac{E}{m}~>~1. $$
Let us (for the fun of it) integrate to find the elapsed proper time $\Delta\tau$ between two radial positions $R_i>R_f$:
$$\begin{align} \Delta\tau~=~&\tau_f-\tau_i~=~ \int_{R_f}^{R_i} \frac{dR}{\sqrt{\left(e+\frac{m}{2R}\right)^2-1}}\cr
~=~& \int_{R_f}^{R_i} \frac{R~dR}{\sqrt{\left(Re+\frac{m}{2}\right)^2-R^2}} \cr
~=~&\left[ \frac{\sqrt{\left(Re+\frac{m}{2}\right)^2-R^2} }{e^2-1} \right.\cr
&-\left. \frac{em\ln\left\{2\sqrt{\left(Re+\frac{m}{2}\right)^2-R^2} + 2R\sqrt{e^2-1}+\frac{em}{\sqrt{e^2-1}} \right\}}{2(e^2-1)^{\frac{3}{2}}}\right]_{R=R_f}^{R=R_i}.\end{align} $$
References:
- E. Poisson, A Relativist's toolkit, 2004; Subsection 3.9, eq. (3.70).
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$^1$ We leave the case of two concentric spherically symmetric thin shells as an exercise for the reader. Obviously, Birkhoff's Theorem comes in handy.