What does the term "Coriolis Effect" refer to? I am familiar with Coriolis Effect referring to the effect that deflects eastward a projectile moving north on the rotating Earth. That is, as something applicable to north/south motion only.
There is also a phenomenon where a projectile shot straight up into the sky will land slightly west of where it was launched, since it was not rotating with the Earth during the time it was in the air.
Recently, I've seen this second phenomenon referred to as "the Coriolis Effect." An example would be this Quora question.
Is this usage correct?
 A: There are usually three fictitious forces that classically arise in a rotating reference frame:


*

*Centrifugal force, which is $m|\Omega|^{2}|r|$ directed away from the axis of rotation.

*Coriolis force, described by $-2m\Omega\times v'$ where $v'$ is the velocity vector of the particle in question as measured in the rotating frame.

*Euler force, which only arises when there is angular acceleration in the rotating frame and is described by $-m\Omega'\times r$ where $\Omega'$ is the time derivative of $\Omega$ and $r$ is the position vector of the particle in the rotating frame.
The term usually referred to as the Coriolis force arises whenever there is a component of velocity perpendicular to the axis of rotation. If a projectile is shot straight up from anywhere except the poles, if a projectile is shot sideways anywhere except the Equator, and if a projectile is shot east or west on the Equator are examples of scenarios where the Coriolis force would introduce some sort of deflection.
To answer your question, yes, both of the phenomena you named can be explained by the Coriolis effect.
A: About firing a projectile straight up into the sky.
At the end of the discussion below I will compare the two cases: north/south projectile motion, and vertical-to-the-local-surface projectile motion.
In the following discussion I'm neglecting air friction.
Here I will use the expression 'angular velocity' to mean: the 'angular velocity of the projectile with respect to the center of the Earth'
As a projectile is fired upward, at the instant of release it has the same angular velocity as the Earth itself. So initally the projectile will move perpendicular to the local surface.
During the flight the motion of the projectile is a Kepler orbit 
The point of highest ascent is the apogee of that orbit.
(Of course, in this case the orbit is extremely low. It's so low that it impacts the Earth in mere seconds. Still, that doesn't change the fact that it is orbital motion.)
As we know, when an object is in orbit then at apogee the angular velocity of that object is at its slowest. As we know, after the apogee the orbiting object is picks up angular velocity again (reaching maximum at perigee).
So: during the ascent the angular velocity of the projectile is decreasing all the time. While initially moving perpendicular to the local surface, the object starts lagging behind the longitude that it was fired from.
After apogee:
During the descent the angular velocity of the projectile increases again. By the time the projectile impacts the surface of the Earth it has regained the starting angular velocity. During all off the trajectory, ascending and descending, there was accumulation of lagging behind.
How much sideways acceleration will there be?
Some definitions:
Radial velocity $v_r$: velocity in the direction of the Earth's center (hence velocity perpendicular to the local surface.
Sideways velocity $v_p$: velocity perpendicular to the radial velocity.
Sideways acceleration $a_p$: acceleration perpendicular to the radial velocity. 
The general convention is to use the greek uppercase Omega, '$\Omega$' for an overall angular velocity that is constant.
Here I will use the lowercase omega '$\omega$' to denote angular velocity because it does change a little. 
We can readily work out what the radial velocity will be as the projectile ascends and descends, what we need is an expression that will get us from $v_r$ to $a_p$ 
The motion of the projectile is orbital motion; as we know during orbital motion angular momentum is conserved.
$$ \frac{d(\omega r^2)}{dt} = 0 $$
Differentiating:
$$ r^2 \frac{d\omega}{dt}  +  \omega \frac{d(r^2)}{dt} = 0 $$
With the chain rule we obtain a term $\frac{dr}{dt}$ which is the $v_r$ part of what we need.
$$ r^2 \frac{d\omega}{dt}  +  2 r \omega \frac{dr}{dt} = 0 $$
Dividing by 'r', and rearranging:
$$ r \frac{d\omega}{dt}  = - 2 \omega \frac{dr}{dt} $$
On the left we have a term $\frac{d\omega}{dt}$; that is angular acceleration. 
If the height that the projectile ascends to is very small compared to the angular velocity then we can treat the $r$ as a constant and move it inside the differentation 
$$ r \frac{d\omega}{dt}  = \frac{dv_p}{dt} $$
In all we have derived this expression from the conservation of angular momentum: 
$$ \frac{dv_p}{dt}  = - 2 \omega \frac{dr}{dt} $$ 
The angular velocity of the projectile changes over time, but compared to the total angular velocity the change is small. 
In all: to a good approximation we can treat the projectile as subject to a sideways acceleration described by the following expression:


*

*$a_p$ acceleration component perpendicular to the radial direction    

*$v_r$ velocity component in radial direction 


$$ a_p = -2\omega v_r $$
In all:  
In the case of east/west deflection of a projectile moving north/south there is no involvement of gravity. The deflection is parallel to the local surface, and gravity is perpendicular to the local surface.
In the case of a projectile shot straight into the sky the deflection goes back to gravity since the Earth's gravity is changing the projectile's angular velocity. Also, as it is gravity that is causing all of the change the force of it is proportional to the mass $m$ of the projectile. 
Both these phenomena are referred to as 'Coriolis effect'
