Coriolis effect discrepancy. How to deduce the direction of the effect?
That means from the perspective of the equator to the north pole, it is rotating clockwise


That's the part you got wrong.
As seen from the Equator (as any other point on the Earth), the North pole is not rotating. The rotation only shows up in an external frame of reference. And is actually using an internal frame of reference (like an observer on the equator) what gives rise to the Coriolis force, that is not a real force but just the inertia seen from a rotating frame of reference.

From the Equator, the North pole is in a fix position, as well as the celestial north pole, that is fix to the Equatorial horizon due north. But the starts and the Sun moves around it from East to West, which is COUNTERCLOCKWISE. That's is because the observer is moving Eastwards.

Coriolis is easier to understood if you imagine that any point on the Earth's surface (or the atmosphere) north of the Equator, have to complete a smaller circle around the Earth axis every 24 hours. Therefore, those points are moving Eastwards at a slower speed than the ones at the equator. Hence, if you shoot a projectile to the North, it will have an excess Eastwards velocity relative to the ground under it, so it will tend to go Eastwards. Which will apear like it is turning rightwards.
If the projectile is flying south from the Equator, Eastwards will mean leftwards.

This schematics from Facts About All will help in an example:


Let's imagine you are at at the monument "La Mitad del Mundo" right on the Equatorial line on Ecuador. And you shoot a projectile straight North to Raleigh, North Carolina, 3,960 km North of "La Mitad  del Mundo".

Let's say your canon shoots a frictionless projectile at speed of 1,000 km/h. Then the projectile had a Northwards speed of 1,000 km/h, and an Eastward speed of 1,666 km/h which is the speed the canon at Ecuador have in order to go around the earth center every 24 hours.

Now, in 3.96 hours, the projectile would have reached the latitude of Raleigh, but in the same amount of time the eastward component of its speed took it 6,600 km to the East. However, the eastward velocity of Raleigh is only 1356 km/h

That is because at latitude 35.7° the distance to the Earth rotation axis is only 5,179 km (in contrast with 6,378 km at the equator), therefore, the distance it have to travel around it every 24 hours is only 32,540 km (in contrast to 40,000) requiring a speed of just 1356 km/h.

That means that, while the projectile wast traveling, Raleigh only moved 5,369 km eastwards, 1,231 km less than the projectile, with the consequence that the projectile ended up, 1,231 km to the East of Raleigh (The right as seen from "La Mitad del Mundo"), catastrophically missing the target, and falling in the Atlantic ocean somewhere north of Bermuda.

This excess eastward velocity visualization of Coriolis, seemingly falls apart when considering a projectile shoot due East or West. But it would still be deflected to the right (in the northern hemisphere). An explanation in such case can be found in this answer, however that answer is not as clear as I would like, as it just ignores the vertical effect, also known as Eötvös effect, which is important to make sense of the deflection of a projectile traveling East/West. In short, the increased/decreased velocity of a projectile moving East/West, increase/decrease the centrifugal force, so it moves away/towards from Earth's axis of rotation. This means "moves up" at the equator, and move south/north and up (in different proportions) north of the equator. The rightwards deflection is considered part of the Coriolis effect, while the upwards deflection is the Eötvös effect.