Why does the Coriolis effect disappear at the equator?

The Coriolis force on the equator indeed does point outwards, if you are moving west to east. This is not the same as the centrifugal force, because the centrifugal force is present always - even if you are not moving. But when you move (west to east), there is an additional force on top of the centrifugal force - the Coriolis force. If you travel east to west on the equator, the Coriolis force is pointing inwards. It has practical effects, as can be seen in this video demonstrating how bullet trajectory is curved upwards or downwards when shooting to the eastern or western direction.

The Coriolis force on the equator disappears if you are traveling in the northern (or southern) direction. In that case your direction is (anti-)parallel to the direction of the $\vec\omega$, so their vector product is zero. Maybe this is what Goldstein had in mind.

According to Wikipedia:

Because the Earth completes only one rotation per day, the Coriolis force is quite small, and its effects generally become noticeable only for motions occurring over large distances and long periods of time, such as large-scale movement of air in the atmosphere or water in the ocean. Such motions are constrained by the surface of the earth, so only the horizontal component of the Coriolis force is generally important.

Since only the horizontal component is usually considered, then both Goldstein's statement and your description can be correct. Goldstein was only considering the non-vertical effects of the Coriolis force.

You are right: the coriolis effect in the equator is converted to a kind of centrifugal force. But in many situations only the horizontal Coriolis forces are accountes, because in the vertical one the gravity is so much greater.

The vertical coriolis and the centrifugal force may point parallel, but they are still different forces. Let's think of a stationary object just above the equator:

From a fixed observer there is no force at all (gravity does not count!).

From a rotating observer, it would look like the object is moving at the same speed as the Earth rotates, but in opposite direction (east to west): $-(\vec\omega\times\vec r)$. It will be the subject of three different apparent forces:

1. Coriolis force: $F_c = 2m(\vec\omega\times(\vec\omega\times\vec r))$, directed downwards.
2. Centrifugal force: $F_r=-m(\vec\omega\times\vec(\vec\omega\times\vec r))$, directed outwards.
3. The apparent course of the object, from the rotational frame of reference, follows a curved path following the shape of the Earth, so the Earthlings will assume an additional centrifugal force to be accounted, identical in magnitude to the former: $F_r=-m(\vec\omega\times\vec(\vec\omega\times\vec r))$.

These three forces added together results in a net force of 0, as expected.