The question you're asking can't be answered using the Schwarzschild metric. This is because the Schwarzschild metric is only valid in static (time-independent, non-rotating) spacetimes (in other words, nothing in the metric can evolve with time). To answer your question you must first generalise the metric from the Schwarzschild solution into the Vaidya metric.
Put another way, if we allow the black hole to absorb matter we must allow the mass parameter to change with coordinate time, this would mean that the metric would no longer yield a timelike Killing vector field. This solution would no longer be stationary (and therefore also not static).
To make this generalisation to our spacetime we take the Schwarzschild solution:
\begin{equation}
ds^2 = -\left(1 - \frac{2M}{r} \right)dt^2 + \left(1 - \frac{2M}{r} \right)^{-1} dr^2 + r^2 d \Omega^2
\end{equation}
and write $t$ in terms of an advanced time coordinate:
\begin{equation}
v = t + r^* \qquad dv = dt + dr^* = dt + \frac{dr}{\left(1 - \frac{2M}{r} \right)} \qquad r^* = r + 2M \log \left(\frac{r}{2M} - 1\right)
\end{equation}
to obtain the advanced Eddington-Finkelstein metric:
\begin{equation}
ds^2 = -\left(1 - \frac{2M}{r} \right) + 2 dr dv + r^2 d \Omega^2
\end{equation}
(note here that if you were instead interested in decreasing mass then we would make the coordinate change $u = t - r^*$)
We then make the change that $M = M(v)$ and we find that this too is a solution to Einstein's equations (which still has spherically symmetry but is no longer stationary). This is known as the (advanced) Vaidya metric and is explicitly given by:
\begin{equation}
ds^2 = -\left(1 - \frac{2M(v)}{r} \right) + 2 dr dv + r^2 d \Omega^2
\end{equation}
This is then the correct metric to use for analysing a spacetime which has a mass that evolves with time. Note here that we have left the evolution of time as a generic function and that it is up to the system under consideration to select the correct behaviour. Generally it is taken to be some linear function.
As for a direct answer to your question, I am having trouble understanding exactly what you're asking.
If you are asking about how radial geodesics will behave then you simply have to use the metric above, just as you would with Schwarzschild homework exercises. An example of this calculation is given in the reference below.
If you're interested in how the horizon will change as a function of time then this will all depend on what kind of functional change the mass $M(v)$ has. In standard exercises it is normally taken to be linear $M(v) = \mu v$.
Remember that due to the spherical symmetry of this solution that the position of a point on the horizon w.r.t. the centre will be given by $p(r,v) r_h(v) = 2M(v)$. It really is as simple as that.
In these lecture notes a lot of details are covered. If you want a more detailed answer then feel free to edit your original post to help me understand exactly what you're interested in. The wikipedia page also covers many of these details.
