It would appear you are asking about the following case: if an object has a velocity with respect to the Earth, in a direction parallel to the local latitude line, will there be a rotation-of-Earth effect?
For this situation let's use an airship. An airship is buoyant, it's propulsion can give it a considerable velocity relative to the Earth, and anything that tends to deflect the motion of the airshil will deflect it, because the airship has no friction with the ground.
An airship that is moving from west to east is circumnavigating the Earth's axis faster than the Earth itself. Compare that to a car going around a curve on a banked circuit.
(Early racing cars had very little grip. On a banked circuit the car is steered up the slope, so that the angle of the slope provides the required centripetal force. From the inside of the bend to the outside the slope is steeper and steeper.)
Conversely, an airship that is moving east to west is circumnavigating the Earth's axis slower than the Earth itself. So the airship will tend to slump down, towards the nearest pole.
About the shape of the Earth:
At the equator the radius of the Earth is 21 kilometers more than the radius at the poles. That is: to go from the equator to the poles is to go downhill. Imagine a planet with exactly the shape of the Earth, but not rotating. Then all fluid would flow to the poles. Over geologic time scale the Earth is deformable enough so that the Earth as a whole is always very close to hydrostatic equilibrium. The shape of the Earth is such that at every latitude the downhill slope is the slope that provides buoyant objects the required centripetal force to remain on the same latitude.
With the above in place, let's calculate how much sideways deflection we can expect, in proportion to velocity (along the local latitude line) with respect to the Earth.
I will use the uppercase letter $\Omega$ for the angular velocity of the Earth, and the lowercase letter $\omega$ for the angular velocity of the airship relative tot the Earth. I will use $\omega_r$ for this relative angular velocity. I will use $R$ for the radius of the Earth.
The expression for the required centripetal acceleration for circumnavigating motion is:
$$ a = \omega^2 r \qquad (1) $$
The airship is moving west-to-east. To find the expected sideways acceleration we calculate the total required centripetal acceleration, and subtract the actually provided centripetal acceleration.
$$ a = (\Omega + \omega_r)^2R - \Omega^2R \qquad (2) $$
$$ a = \Omega^2R + 2 \Omega \omega_rR + \omega_r^2R - \Omega^2R \qquad (3) $$
$$ a = 2\Omega \omega_rR + \omega_r^2R \qquad (4) $$
Now we convert the relative angular velocity of the airship to a relative linear velocity ( $v=\omega r$ )
$$ a = 2\Omega v_r + v_r^2/R \qquad (5) $$
For the kind of velocity that an airship can reach the first term, $2\Omega v_r$, is much larger than the second term $v_r^2/R$
So:
Unless the crew of the airship takes countermeasures an airship moving along a local latitude line, with a velocity with respect to the Earth, will veer away from that course. The tendency to veer away is proportional to the velocity with respect to the Earth. The magnitude of the tendency to accelerate sideways will be $2\Omega v_r$. The direction of the veering is as follows: an airship moving west-to-east will swing wide, veering to the outside of the local latitude line. An airship moving east-to-west will veer to the inside of the local latitude line.
Note that for this calculation it isn't necessary to know the mass of the airship. The required centripetal acceleration is provided by the Earth's gravity, and inertial mass and gravitational mass are equivalent.
Also, at any latitude the direction of the tendency to deflect is parallel to the plane of the equator. But at higher latitudes the Earth surface is at an angle to the plane of the equator. A local calculation must take that into account.
For an airship:
Maintaining altitude is managed by adjusting the buoyancy of the airship.
Maintaining course is managed by pointing the nose of the airship in the direction opposite to the tendency to veer away from the intended course.
In itself the rotation-of-Earth effect is a single effect, the direction is in the plane of the rotation. It's just that when you are flying an actual airship altitude and course are managed separately. So it's practical to think of the total rotation-of-Earth effect as decomposed in a vertical-to-the-local component and a horizontal-to-the-local component.
Meteorologists refer to the horizontal-to-the-local component as 'Coriolis effect'.
