In general, at every point, the acceleration of the particle has tangential as well as radial components, just as it has both x and y components. So $\ddot x$ is the sum of the x components of both radial (ie centripetal) acceleration $v^2/R$ and tangential acceleration. You have missed out the x component of tangential acceleration.

In general, at every point, the acceleration of the particle has tangential as well as radial components, just as it has both x and y components. So $\ddot x$ is the sum of the x components of both radial (ie centripetal) acceleration $\frac{v^2}{R}$ and tangential acceleration. You have missed out the x component of tangential acceleration.

At A the y-component of acceleration makes no contribution to centripetal acceleration, so the x-component of acceleration equals the centripetal acceleration - ie $a_x=\frac{v_0^2}{R}$. The x-component of velocity at A is $\dot x=u_x=0$.

At B the particle has no y-component of velocity $(v_y=\dot y=0)$, only an x-component $v_x$. It has travelled a distance $R$ horizontally so
$v_x^2=u_x^2+2a_x R$
from which the velocity $v=v_x$ at B can be found.