Firstly, is that correct?
Yes your intuitive understanding for this part of the Coriolis effect is correct.
The second part, that is, why wind in the East direction is deflected South, is a bit trickier, and involves the use of centripetal force. this is given by the equation:
$F = \frac{mv^2}{r}$
If we re-arrange the above equation, we can find $r$ in terms of $v$, and we arrive at:
$r = \frac{mv^2}{F}$
This tells us that as velocity increases, the radius required to maintain the orbit also increases.
Now let's apply this concept to winds on the Earth. If we feel no wind on the Earth, then the air in the atmosphere is travelling at the same velocity as the Earth. The Earth is naturally spinning towards the East.
In the case of an additional Eastward wind felt on the Earth, this wind has effectively increased its velocity, and therefore the above equation tells us that the radius of orbit must increase as well. Radius in this case is the distance, measured perpendicularly of the Earth's axis, between the axis and the wind.
In order for the radius to increase, the wind moves southwards, where the radius is larger.
Similarly, wind moving in the West direction, is moving in the direction opposite of that to the Earth, and therefore its velocity is decreased. Consequently this wind moves towards the North, where the radius is less.
The above image shows what happens. The wind moving East begins to expand its radius, thus moving outwards. Gravity pulls it back, and the wind moves South, in order to maintain the larger radius required for its increased velocity.
Coriolis force is not an actual force, but rather an effect observed in rotating frame of reference. The light path is not actually bent, so it doesn't matter that the photon has no mass, the Earth's rotation will have an affect on the photon's apparent path.
This does not contradict your calculation of $F_{Coriolis}=0$, because you have to put this force in $F=ma$, where $m$ is, again, zero. However, you can take the limit of both sides for $m\rightarrow 0$ and calculate $a$.
Best Answer
The horizontal component of the Coriolis force for non-equatorial latitudes arises because the axes of the rotating frame are not parallel to the axis of rotation, except for when the rotating frame is on the equator.
The statement in the question, that $\Omega$ is parallel to $v'$, is incorrect.
You can see that if an object is traveling North, there is a component of the angular velocity that is perpendicular to the object's velocity. The right hand rule gives $\Omega \times v'$ pointing in the +y' direction, but there is a negative sign on the expression for the Coriolis force, so the horizontal component of the Coriolis force points in the -y' direction. This corresponds to East, which is the correct direction.
The magnitude of the Coriolis force can be found using the trigonometric breakdown shown below.
