You made a few errors in transcribing the standard formula
$$\arccos\left(\frac{x_1x_2+y_1y_2+z_1z_2}{\sqrt{x_1^2+y_1^2+z_1^2}\sqrt{x_2^2+y_2^2+z_2^2}}\right). $$
But your suspicion that it was not applicable is correct.
This is the formula for the angle between the vectors
$(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$,
that is, it tells you, if you were sitting at the point $(0,0,0)$
watching the projectile through a telescope,
how much you would need to turn the telescope in order to keep it pointed
at the projectile.
If the positions $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$
are not too close to $(0,0,0)$ and are observed very close in time,
then naturally you will not have to turn the telescope very much in order to track the projectile.
To find the direction of travel your projectile, a more useful vector would be the vector from one observed point to the next observed point,
$$ (x_2 - x_1, y_2 - y_1, z_2 - z_1). $$
This vector is the sum of a horizontal vector $(x_2 - x_1, y_2 - y_1, 0)$
and a vertical vector $(0,0, z_2 - z_1)$.
The vector sum can be drawn graphically as a right triangle with those two vectors as the legs and the vector $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$ as the hypotenuse.
The length of the hypotenuse is therefore
$$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} $$
and the length of the vertical leg is $z_2 - z_1$, so by simple trigonometry
the angle from straight-up vertical is
$$ \arccos\left(\frac{z_2 - z_1}
{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}\right) $$
and the angle from the horizontal is
$$ \arcsin\left(\frac{z_2 - z_1}
{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}\right). $$
But since the projectile travels in a curve rather than a straight line,
the angle of this vector is only a kind of "average" angle of travel over the arc from the first point to the second point, not the actual angle at either endpoint.
Assuming a simple parabolic arc for the path of the projectile,
the angle you get from the formulas above is the exact angle of travel at the point
$$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} + \zeta\right)$$
for some $\zeta> 0,$ that is, at some point whose horizontal coordinates are midway between the two observed points, but whose vertical position is somewhere above the line connecting the two points.
How far above the line that position will be is a function of the acceleration of gravity and of how much time passed between the observations.
If you know the acceleration of gravity and the time that passed between observations then you can work backward from the angle measured in the above formulas to the angle at the point $(x_1,y_1,z_1)$.
But a simpler method for a parabolic arc is that if you want the direction of travel at a particular time $t$, don't use the position at time $t$; instead, use the positions at time $t - \delta$ and at $t + \delta$,
because then the point at which the slope of the arc matches the slope of the vector will be the point at time $t.$
For the horizontal direction, you have the right idea about looking at the points $(x_1,y_1)$ and $(x_2,y_2)$ in the $x,y$ plane.
It's also true that the slope of the line between those points is not an angle in the usual meaning of the word "angle."
In particular, I think you would want to distinguish a trajectory that passes through $(x_1,y_1)$ first and later through $(x_2,y_2)$ from a trajectory that passes through $(x_2,y_2)$ first and then $(x_1,y_1)$;
but the formula $(y_2 - y_1)/(x_2 - x_1)$ gives the same answer in both cases.
For the horizontal direction I suggest you consider the direction of the vector $(x_2 - x_1, y_2 - y_1)$ in the $x,y$ plane.
There is some discussion of the angular direction of this vector in
How do we really get the angle of a vector from the components?.
To summarize, the angle is essentially the result of the inverse tangent
(aka arc tangent) function.
