First a review of how to determine the direction of the supports
In these cases, it's always useful to imagine what would happen if there were no supports. In this case, what would happen if there were no bolts.
As you noticed, one thing that'd happen is that the piece would rotate counter-clockwise. We don't want that to happen, so we can already tell that the supports will need to resist that rotation.
But the piece would obviously also move to the left. So the supports will need to resist that motion as well.
In general, we want to satisfy three conditions for any 2D object:
$$\begin{align}
\sum F_x &= 0 \\
\sum F_y &= 0 \\
\sum M_0 &= 0
\end{align}$$
That is, the sum of external forces (including reactions) in the horizontal direction must be zero, as must the sum of vertical forces and the sum of moments around any arbitrary point.
Now, let's start with the first equation. Subscripts 1 and 2 describe the two screws, and I start by assuming both reactions point to the right, and are therefore positive:
$$\begin{align}
\sum F_x &= R_{1,x} + R_{2,x} - F = 0 \\
\therefore F &= R_{1,x} + R_{2,x}
\end{align}$$
This is pretty obvious: the two reactions must cancel out the applied horizontal force. But how the screws divide that load is still unknown (maybe they split it 50-50, maybe the top screw absorbs it all by itself, maybe the bottom one has a reaction greater than the applied force and the top one then has to cancel out that excess reaction... we don't know yet).
Now for the second equation (now assuming both forces point up and are positive):
$$\begin{align}
\sum F_y &= R_{1,y} + R_{2,y} = 0 \\
\therefore R_{1,y} &= -R_{2,y}
\end{align}$$
The applied force has no vertical component, so we only need to balance out the vertical components of the reactions. The difference in sign tells us they must be equal and opposite: maybe they're both zero, maybe they're a billion tons each (in different directions). Still to be determined.
Now for the last equation, calculating the moment around the bottom screw (remembering that moment is equal to force times perpendicular distance from the force to the selected point). I'll adopt $d$ as the distance between the screws and $L$ as the vertical distance from the force to the bottom screw. Forces which would cause a counter-clockwise rotation are positive.
$$\begin{align}
\sum M_2 &= -R_{1,x} \cdot d + R_{2,x} \cdot 0 + R_{1,y} \cdot 0 + R_{2,y} \cdot 0 + F \cdot L = 0 \\
-R_{1,x} \cdot d + F \cdot L &= 0 \\
\therefore R_{1,x} = \dfrac{F \cdot L}{d}
\end{align}$$
Notice that $R_{2,x}$ and $R_{2,y}$ have a distance of zero and therefore are discarded for this calculation. This makes sense, since a force on a point doesn't cause rotation around that point. Likewise, $R_{1,y}$ also has zero distance since we care about the perpendicular distance, and that force's "line of action" (it's projection) crosses through the other screw.
So we now know the direction of the force on the top screw: the opposite of what you drew, it will point to the right, countering the applied force.
Using $\sum F_x$, we can then see that
$$\begin{align}
R_{2,x} &= F - \dfrac{F \cdot L}{d} \\
&= F\left(1 - \dfrac{L}{d}\right)
\end{align}$$
As drawn, $L$ is obviously greater than $d$, which would make this force "negative", meaning our initial hypothesis that it was pointing to the right was incorrect: it actually points to the left, the same direction as $F$. This makes sense: the previous calculation told us the top screw's reaction was greater than the applied force, so the bottom screw now needs to compensate for that excess reaction.
And now to answer your actual question
Notice that the entire calculation above didn't take the structure's shape into consideration in any way. Whether the beam was a diagonal or a U-shape or an M.C. Escher painting was irrelevant. No matter the shape of your beam, or whether you add more beams, the reaction will always be the same.
You can even add more screws between the existing ones and the conclusion won't change: more screws will reduce the load on each screw, but you'll still have some screws reacting to the right and others to the left.
But notice the caveat in that previous paragraph: "You can even add more screws between the existing ones". That's a hint to our eventual solution. And that hint points us back to the equation for $R_{1,x}$:
$$R_{1,x} = \dfrac{F \cdot L}{d}$$
As mentioned, the screws have reactions in different directions because $R_{1,x}$ is actually greater than $F$. So, how can we fix that?
Well, we just need to increase the spacing $d$ between the screws, such that it is greater than $L$. If that happens, then $R_{1,x} < F$, therefore
$$R_{2,x} = F\left(1 - \dfrac{L}{d}\right) > 0$$
If $R_{2,x}$ is greater than zero, that means our initial hypothesis that it also points to the right was correct.
