If you project the velocity of a water parcel traveling up the channel into radial coordinates in both the inertial frame (space frame) and the rotating frame of reference (earth frame) then I think the necessary effects will become apparent.
In the earth's frame, the velocity is (use a ' to symbolize the rotating frame):
$$\vec{v} = 0\cdot\hat{r}' + v_{flow}\cdot\hat{\theta}' + 0\cdot\hat{\phi}'$$
In the space frame, the velocity is:
$$\vec{v} = 0\cdot\hat{r} + v_{flow}\cdot\hat{\theta} + v_{\theta \:max}\cdot\cos\theta\cdot\hat{\phi}$$
For both frames: If the water flows at a constant rate in the theta direction, then the theta component of the velocity vector is constant. Also, the r component will be zero for both cases since we assume the earth radius to be constant.
For the earth frame:
If the water is flowing directly from south to north, then the phi component is zero in the earth's rotating frame.
For the rotating frame:
The water flowing from south to north means it must be picking up speed in the $\hat{\phi}$ direction. This is kept track of by the $v_{\theta \:max}\cdot\cos\theta\cdot\hat{\phi}$ in the space frame velocity equation.
In the nonrotating frame, the change in velocity in the $\hat{\phi}$ direction means there must be an acceleration to produce it. This means the water must get an acceleration from the west wall toward the phi direction while in the southern hemisphere in order to increase the velocity enough to keep up with the earth. This will happen until you cross the equator which at that point the water has to start to slow down to keep tracking in the channel on the earth. This means there must be an acceleration, but this time from the east wall of the channel.
Long story short, the west wall will erode in the southern hemisphere and the east wall will erode in the northern hemisphere. You can do more math by taking derivatives to get the acceleration as a function of angle and then find the spots of greatest erosion, etc.
