Alfred wrote: "An observer which is stationary in a Schwarzschild metric [...] drops a particle with initial zero velocity"
If the observer is stationary his coordinate acceleration is
$$\rm \frac{d^2 x^{\mu}}{d \tau^2}=\{ \ddot{t}, \ \ddot{r}, \ \ddot{\theta}, \ \ddot{\phi}\}=\{0,\ 0, \ 0, \ 0\}$$
his four-velocity
$$\rm \frac{d x^{\mu}}{d \tau}=u^{\mu}=\{ \dot{t}, \ \dot{r}, \ \dot{\theta}, \ \dot{\phi}\}=\{ 1/\surd g_{tt}, \ 0, \ 0, \ 0\}$$
and his four-acceleration
$$\rm a^{\mu}= \frac{d^2 x^{\mu}}{d \tau^2}+\sum_{\alpha, \sigma} \ \Gamma^{\mu}_{\alpha \sigma} \ u^{\alpha}u^{\sigma}$$
so the proper acceleration he reads on his accelerometer is
$$\rm |a|=\frac{F}{m}= \surd \ |\sum_{\mu, \nu} g_{\mu \nu} \ a^{\mu} a^{\nu}| =\frac{G M}{r^2 \sqrt{1-r_s/r}}$$
That is also the acceleration of the particle he just dropped in his frame, although in terms of the noneuclidean coordinates where $\rm r$ has depth expansion the $\rm d^2r/dT^2$ is weaker by a factor of $\rm 1/\surd g_{rr}$ (that coincidentally cancels the $\rm \surd g_{tt}$ from the proper acceleration in that direction), which in the $\rm v\to 0$ limit gives $\rm d^2 r/dT^2=-G M/r^2$ (see below).
Alfred asked: "So my question is, what are the vertical and horizontal components of this acceleration in my observer's referential of a particle at the exact time it passes the observer with a nonzero velocity?"
In your example where the particle with the local three-velocity $\rm v$ is in free fall his $\rm a=0$, so only the radial $\rm a$ of the stationary observer remains.
The cartesian $\rm {d^2 x^{\mu}}/{d T^2}=\sum_{\alpha, \sigma} \ \Gamma^{\mu}_{\alpha \sigma} \ u^{\alpha} \ u^{\sigma}/ \gamma^2$ of the free falling particle in the frame of the local stationary observer is, with $\rm T$ for the local shell time, $\rm r^2=X^2+Y^2+Z^2$ and $\rm \beta=v/c$:
$\rm \frac{d^2 X}{d T^2}=\frac{c^2 \ r_s \ X \left(-r^3+r^2 \ r_s+\beta ^2 \left(r \left(X^2-2 \left(Y^2+Z^2\right)\right)+2 r_s \left(Y^2+Z^2\right)\right)-r_s \ X^2\right)}{2 r^3 \left(r^3-r^2 \ r_s+r_s \ X^2\right)}$
$\rm \frac{d^2 Y}{d T^2}=\frac{c^2 \ r_s \ Y \left(\left(2 \beta ^2+1\right) \left(Y^2+Z^2\right) \left(-r \ r_s+Y^2+Z^2\right)-\left(\beta ^2-1\right) X^4+\left(\beta ^2+2\right) X^2 \left(Y^2+Z^2\right)\right)}{2 r^4 \left(r_s \left(Y^2+Z^2\right)-r^3\right)}$
$\rm \frac{d^2 Z}{d T^2}=\frac{c^2 \ r_s \ Z \left(\left(2 \beta ^2+1\right) \left(Y^2+Z^2\right) \left(-r \ r_s+Y^2+Z^2\right)-\left(\beta ^2-1\right) X^4+\left(\beta ^2+2\right) X^2 \left(Y^2+Z^2\right)\right)}{2 r^4 \left(r_s \left(Y^2+Z^2\right)-r^3\right)}$
where $\beta$ is in the $\rm X$ direction. For a radial trajectory set $\rm X=r$, for a tangential trajectory $\rm Y=r$ or $\rm Z=r$ and for arbitrary directions relative to the stationary observer choose arbitrary position coordinates for the place where the free faller flies past the stationary observer.
In the limit $\rm v\to c$ a horizontal ($\rm Z=r$) photon has a vertical (in the $\rm Z$ direction) acceleration of $\rm 2 \beta^2+1=$$ \ 3 \times$ the naively expected $\rm -G M/r^2$ (and therefore twice the deflection angle when integrated over a long distance), and a vertical ($\rm X=r$) photon $1/\gamma^2=$$ \ 0 \times$ since $\rm c$ is already the maximal local velocity (for an example with numbers see here).
The radial coordinate in Schwarzschild is depth expanded though, so multiply the radial component of the $\rm {d^2 x^{\mu}}/{d T^2}$ vector with $\rm \surd g_{rr}$ to transform to local flat normal coordinates. This restores the expected proper acceleration of the stationary observer in the $\rm v\to 0 $ limit:
$$\rm \left\{\frac{d^2 \bar{X}}{d T^2} , \ \frac{d^2 \bar{Y}}{d T^2}, \ \frac{d^2 \bar{Z}}{d T^2} \right\} = \left\{\frac{d^2 X}{d T^2} \ \surd g_{{}_{XX}} , \ \frac{d^2 Y}{d T^2} \ \surd g_{{}_{YY}} , \ \frac{d^2 Z}{d T^2} \ \surd
g_{{}_{ZZ}} \right\}$$
where $\rm g_{{}_{XX}}=1+\frac{r_s \ X^2}{(r-r_s) \ r^2} , \ g_{{}_{YY}}=1+\frac{r_s \ Y^2}{(r-r_s) \ r^2} , \ g_{{}_{ZZ}}=1+\frac{r_s \ Z^2}{(r-r_s) \ r^2}$.
To transform the stationary observer's proper acceleration vector $\rm \vec{a}$ into his local kinematic acceleration $\vec{a} $ in the free falling particle's frame the transformation is:
$$\vec{a}={\rm \frac{1}{\gamma^{2}}\left[\vec{a}-\frac{(\vec{a}\cdot\vec{\beta}) \ \vec{\beta}}{|\beta|^{2}}\left(1-\frac{1}{\gamma}\right)\right]}, \ \ \vec{\rm a}= \gamma^{2} \left[\vec{a}+\frac{(\vec{a}\cdot\vec{\beta}) \ \vec{\beta}}{|\beta|^{2}}\left(\gamma-1\right)\right]$$
(for an example with numbers see here). If the particle's velocity is purely longitudinal or tangential to the proper acceleration of the observer divide the longitudinal component of $\rm a$ by $\gamma^3$ or the tangential components by $\gamma^2$.
Locally we have the equivalence principle and Minkowski, so the special relativistic equations hold at the moment when they fly past each other.
The proper time derivative of a free falling observer's radial velocity measured relative from one stationary point to the next (not relative to a single point) is in his own frame
$$\rm \dot{v}=\frac{dv}{d\tau}=\frac{c^3 \left(2 \ \ddot{r} \ r \ (r-r_s)-\dot{r}^2 \ r_s\right)}{2 \ \sqrt{r} \ \left(c^2 \ (r-r_s)+\dot{r}^2 \ r\right)^{3/2}}$$
where $\rm \dot{r}=v \ E, \ \ddot{r}=-G M/r^2, \ E=\sqrt{(1-r_s/r)/(1-|\beta|^2)}$ (the latter is constant). In the special case where the particle is falling from infinity or escaping with the escape velocity $\rm v=c \ \sqrt{r_s/r}$ we have $\rm E=1, \ v=\dot{r}$ and the equation reduces to $\rm \dot{v}=\ddot{r}=-G M/r^2$.
If the particle also has tangential velocity this becomes a little bit less elegant, but since the question wasn't about the $\rm dv/d\tau$ I just brought that up for comparison anyways, in your example we only need the $\rm a$ or $\rm g$ which you already had and $\rm \gamma=\dot{t} \surd g_{tt}=1/\sqrt{1-|\beta|^2}$ to do the local Minkowski transformation between proper and kinematic acceleration.
There is a related answer for particles with tangential velocity by the way, see here. Doublecheck my answer though to make sure there aren't any typos in the Latex or errors in my thinking. I would also like to hear a 2nd opinion, which I would upvote as well.
