Why do hydrostatic fluid exert force on any wall normally with magnitude ρgh always

Imagine a piston of small area, A inside an empty cylinder, close to its bottom end, which is open. The cylinder is immersed in liquid, at any angle, so that the piston is at depth h and orientated at any angle. When the piston is pushed down the cylinder by a small distance $Delta x$, let's say a force $F_bot$ normal to the piston is needed. Thus the work done by the piston on the liquid is $F_bot Delta x$. But the displaced liquid (volume $A Delta x$) makes the liquid level rise; effectively the displaced liquid rises by height $h$ and gains gravitational potential energy $mgh = rho A Delta x gh.$ If the fluid is perfectly non-viscous, or the process is very slow (quasi-static), we can equate the work done to this gain in PE. Thus: $$F_bot Delta x=rho A Delta x gh.$$ Dividing both sides of the equation by $A Delta x$ and writing $p=frac{F_bot}{A}$ we get $$p=h rho g.$$

Once you've seen the power of this argument, you'll see that it works anywhere in the liquid, including at a wall of the container. Indeed you could make a small part of the wall the movable piston itself.

I assumed that since very small volume of liquid is at rest, it experiences equal and opposite force from both the liquid side and wall side, but from the liquid side in the normal direction to the wall pressure is found to be 2ρgh {integrating all the normal components of pressure},

You do not need to do integration here. Let's draw a free body diagram, and lets assume that the small volume is so small that we don't need to worry about pressure changes from top to bottom but the pressure is $rho g h$ everywhere.

Now, let's calculate $vec f_1$, $vec f_2$, and $vec f_3$ using $vec f=P vec A$ directly since those are in the fluid and we know $P=rho g h$. Then by Newton's 2nd law $vec f_0=-(vec f_1+vec f_2+vec f_3)$. So we have $$vec f_1=rho g h (0,d^2)$$ $$vec f_2=rho g h (d^2, 0)$$ $$vec f_3=rho g h left(0,- d left[d-d tan(theta) right] right)$$ and therefore $$vec f_0=-(vec f_1+vec f_2+vec f_3)=left(-rho g h d^2, -rho g h d^2 tan(theta)right) $$

So $$||vec f_0||=f_0=rho g h d^2 sec(theta)$$ $$hat f_0 = vec f_0/f_0=left(-cos(theta),-sin(theta)right)$$ And since $A_0 = d^2/cos(theta)$ we find that the force is directed normal to the area and has a magnitude of $rho g h A_0$, so $f_0= P vec A_0$ as expected, even at a slanted wall.