Well, the most general circular motion can be described by an angle ##\phi(t)##. Let the circle be in the origin of the ##xy## plane. Then the trajectory is given by
$$\vec{x}(t)=R \begin{pmatrix} \cos[\phi(t)] \\ \sin [\phi(t)] \end{pmatrix}.$$
Now you have to take the 1st and 2nd time derivatives to get velocity and acceleration:
$$\vec{v}(t)=\dot{\vec{x}}(t)=R \dot{\phi}(t) \begin{pmatrix} -\sin[\phi(t)] \\ \cos[\phi(t)] \end{pmatrix},$$
$$\vec{a}(t)=\dot{\vec{v}}(t)=\ddot{\vec{x}}(t) = R \ddot{\phi}(t) \begin{pmatrix} -\sin[\phi(t)] \\ \cos[\phi(t)] \end{pmatrix}-R \dot{\phi}^2(t) \begin{pmatrix} \cos[\phi(t)] \\ \sin [\phi(t)] \end{pmatrix}.$$
As you see, the velocity is (as for any motion) always pointing along the tangent of the trajectory. The acceleration splits into two parts: The tangential acceleration of magnitude (and sign wrt. the direction of the tangent vector) ##a_{\parallel}=R \ddot{\phi}## and one perpendicular, i.e., along the position vector. The component is ##a_{\perp}=-R \dot{\phi}^2 \leq 0##, which means it's always negative, i.e., directed towards the center. The prependicular component is called centripetal acceleration.
According to Newton's Law to maintain this motion you need the total force
$$\vec{F}=m \vec{a}.$$
The part in direction perpendicular to the trajectory is called centripetal force.