As
@jbriggs444 pointed out, the key relation that links the position of an off-center point P ##\vec{r}##, inear velocity of point P ##\vec{v}## and the angular velocity ##\vec{\omega}## about the center of the wheel O is $$\vec{v}=\vec{\omega}\times\vec{r}.$$Now consider a wheel rolling to the right in the plane of the screen and a right-handed coordinate system such that
##\hat x =~## unit vector to the right
##\hat y =~## unit vector down (from the center to the point of contact)
##\hat z =~## unit vector into the screen
Let point P be at the 12 o'clock position on the rim at distance ##R##. Clearly, ##\vec{r}=-R\hat y##. Then $$\vec{v}_P=\vec{\omega}\times\vec{r}_P=\omega\hat{k}\times(-R\hat y)=-\omega R(\hat k\times \hat y)=+\omega R(\hat y\times \hat k)=\omega R \hat{x}.$$This says that point P moves to the right relative to the center of the wheel.
Now let point Q be at the 6 o'clock position at distance ##R##. Clearly, its position vector is the negative of the position vector of P. Then $$\vec{v}_Q=\vec{\omega}\times\vec{r}_Q=\vec{\omega}\times(-\vec{r_P})=-\vec{\omega}\times\vec{r_P}=-\vec{v}_P.$$ The 12 o' clock point is moving to the right whilst the 6 o' clock position is moving to the left. That's a clockwise rotation.
For the rolling (without slipping) part of the motion you shift the reference point to the point of contact Q ##(\vec{r}_Q=0.)## You can easily show by the same method that
##\vec{v}_Q=0##, the point of contact is instantaneously at rest.
##\vec{v}_O=\omega R \hat{x}##, the center moves to the right with speed ##\omega R.##
##\vec{v}_P=2\omega R \hat{x}##, point P moves to the right with speed ##2\omega R.##