What is the coordinate acceleration for pure radial motion?
As Rob Jeffries suggested in a comment, formulating a solution for arbitrary dr/dt requires another parameter, either the energy per unit mass, E/m, or the radius of the shell where the object was dropped.

A shell observer at $r_0$, sees an object passing by at proper radial velocity $v_0$, with corresponding energy/momentum as measured in the local Minkowski frame (using units with c = G = 1)
$$m^2=E_0^2-p_0^2=E_0^2-\gamma _0^2 m^2 v_0^2$$
$$\frac{E_0^2}{m^2}=\gamma _0^2 v_0^2+1=\frac{v_0^2}{1-v_0^2}+1=\frac{1}{1-v_0^2}=\gamma _0^2$$
where the subscripts indicate local shell coordinates. In Schwarzschild coordinates,  energy per unit mass is constant and defined as
$$\frac{E}{m}={ \left(1-\frac{2 M}{r}\right)}\frac{dt}{d\tau}$$
From equations 11 and 12 in section 6.3 of the Taylor, Wheeler, Bertschinger book (TWB) linked in the original question
$$\frac{E_0}{m}=\left(1-\frac{2M}{r_0}\right)^{-1/2}\frac{dt}{d\tau}$$
$$\frac{E}{m}=\sqrt{1-\frac{2M}{r_0}}\frac{E_0}{m}=\gamma _0 \sqrt{1-\frac{2 M}{r_0}}=\sqrt{\frac{1-\frac{2M}{r_0}}{1-v_0^2}}$$
The equation for direct radial inward motion is obtained from TWB section 8.4, equation 17, by setting the angular momentum to zero and taking the negative root .
$$ \left(\frac{dr}{d\tau}\right)^2=(E/m)^2-\left(1-\frac{2M}{r}\right)$$
$$\frac{dr}{dt}=\frac{dr}{d\tau}\frac{d\tau}{dt}= 
\frac{1-\frac{2 M}{r}}{E/m}\frac{dr}{d\tau}$$
$$\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\sqrt{1-\frac{1-\frac{2M}{r}}{(E/m)^2}}$$
Differentiating, and a bit of algebra yields the following result for the acceleration.
$$\frac{d^2 r}{dt^2}=-\frac{2M}{r^2}\left(1-\frac{2 M}{r}\right)
\left(1-\frac{3}{2}\frac{1-\frac{2 M}{r}}{(E/m)^2}\right)$$
These can be rewritten in terms of the Schwarzschild radius $r_s=2M$ and proper velocity, $v_0$, at $r_0$.
$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-\frac{1-\frac{r_s}{r}}{(E/m)^2}}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}}$$

$$\frac{d^2 r}{dt^2}=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right)
\left(1-\frac{3}{2}\frac{1-\frac{r_s}{r}}{(E/m)^2}\right)
=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right)
\left(1-\frac{3}{2}(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}\right)$$

For known $dr/dt$ at a radius $r$, $E/m$ can be determined by inverting the previous expression for $dr/dt$. Then substitute $E/m$ in the equation for acceleration above.

$$(E/m)^2=\frac{(1-r_s/r)^3}{(1-r_s/r)^2-\left(\frac{dr}{dt}\right)^2}$$

The alternate approach, dropping an object from rest at radius $r_0$, uses equations 37 and 38 from section 6.7 of TWB, resulting in

$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)
\sqrt{\frac{\frac{r_s}{r}-\frac{r_s}{r_0}}{1-\frac{r_s}{r_0}}}$$

$$\frac{d^2 r}{dt^2}=\frac{r_s\left(1-\frac{r_s}{r}\right)}
{2 r^2\left(1-\frac{r_s}{r_0}\right)}
\left(1+\frac{2r_s}{r_0}-\frac{3r_s}{r}\right)$$

The shell radius $r_0$ necessary to obtain a desired velocity at shell $r_1$ can be found using
$$\frac{E_0}{E_1}=\sqrt{\frac{1-\frac{r_s}{r_1}}{1-\frac{r_s}{r_0}}}
=\frac{\gamma_0}{\gamma_1}$$
where $\gamma_0=1$ for an object dropped from rest at $r_0$.

As an example, a plot is shown below using $v_1=1/2$ at $r_1=2r_s$. For these values $\gamma_1=2/\sqrt{3}$, $E/m=\sqrt{2/3}$ and $r_0=3r_s$. For reference, the green and red curves are Schwarzchild coordinate velocity and acceleration for an object dropped from rest at infinity, $E/m=1$. The dashed blue and brown curves are for an object dropped from rest at $r_0$, and the solid blue and brown curves for an object launched from $r_1$ with a Schwarzschild velocity of
$$\left(\frac{dr}{dt}\right)_{r_1}
=-\left(1-\frac{r_s}{r_1}\right)v_1
=\frac{1}{4}$$
Note that the dashed curves are overlaid by the solid ones from $r_1$ to $r_s$, since the initial conditions were chosen to result in the same velocity at $r_1$.