I think you are confusing the thrower with the observer.
In the left picture, the observer is in the middle of the carousel but not rotating with it. You can imagine that the observer is hovering a bit over the center of the carousel.
In the right picture, the observer is also in the middle of the carousel, but he is now attached to it, so he’s rotating relative to the ground. To this observer the ball’s trajectory will indeed appear to be curving to the right.
You are essentially correct however: from the point of view of the thrower, the path of the ball will appear as bending the other way, to his left.
Edit: On a closer look, and following our discussion in the comments, I think that when they write:
Kinematics insists that a force (pushing to the right of the instantaneous direction of travel for a counter-clockwise rotation) must be present to cause this curvature
They are referring to the point of view of the thrown object. This is because that’s the best way to make sense of the mention of “instantaneous direction of travel”. I think they are referring to the instantaneous direction of travel of the thrown ball. Then it makes complete sense because relative to the carousel it really is bending to the right all throughout its motion, from the edge of the carousel towards its center (and in fact also after that).
Just for the sake of clarity, I am adding a 180 degrees rotated version of the right diagram you have attached, showing how in the rotating frame, the ball’s trajectory curves in fact to the right relative to the direction of its own instantaneous direction of motion:
The important thing to notice here, is that the motion that the thrower will observe is a combination of two fictitious forces:
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The centrifugal force, is what initially makes the ball appear to veer a bit to the left side of the thrower. That’s because it is the centrifugal force that gives objects in a rotating frame the tendency to move away from the center of rotation and towards the periphery, “throwing stuff out”, as it were.
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The Coriolis force, is what is responsible for the clockwise motion of the ball. You have deduced correctly in the comments, by applying the right hand rule, that the Coriolis force acting on the ball points to its right. Meaning, $90^o$ clockwise relative to the ball’s instantaneous velocity.
Also, we can determine the direction of the Coriolis force quite easily:
In the rotating frame, we have: $ \boldsymbol{\vec{v}_0} = -v_0\boldsymbol{\hat{y}}$, as the initial velocity is down towards the center when the ball is at the top, as in the right diagram you attached.
For the angular velocity relative to the ground, we have counterclockwise motion so $\boldsymbol{\vec{\omega}} = \omega\boldsymbol{\hat{z}}$, (with $\omega>0$). This gives:
$$ \boldsymbol{\vec{F}_{\text{coriolis}}} = -2m\ \boldsymbol{\vec{\omega}}\times\boldsymbol{\vec{v}_0} = -2m\begin{vmatrix} \boldsymbol{\hat{x}} & \boldsymbol{\hat{y}} & \boldsymbol{\hat{z}} \\ 0 & 0 & \omega \\ 0 & -v_0 & 0 \end{vmatrix} = -2m\omega v_0\boldsymbol{\hat{x}} $$
This makes sense: the force is acting towards the negative $x$ axis, but the ball is thrown from the other side, so the resulting force on the ball is to the right relative to the direction of the throw.
This is a well known result that in a counterclockwise rotating frame, the Coriolis force induces clockwise motion (and vice versa).
Take note however that $\boldsymbol{\vec{v}_0}$ is measured in the rotating frame, so it is not constant. This entire computation is good for the initial velocity. An instant after the throw, the velocity vector will begin deviating from pointing radially inwards in the rotating frame.
