You need to discern material points and geometric points:

• let's call $T$ the geometric point identifying the top of the disc at all the times;
• let' call $P$ a material point, s.t. $P \equiv T$
• let's call $D$ a material point on the surface of the disc.

Here some facts will show below:

• The geometric point $T$ (and the material point $P \equiv T$) always has velocity $\mathbf{v}_T = \mathbf{v}_P = \Omega R \mathbf{\hat{x}}$;
• the material point $D$ has velocity $\mathbf{v}_D = 2 \Omega R \mathbf{\hat{x}}$ only when it passes through $T$;
• all the material points, when they pass through $T$, always has velocity $\mathbf{v} = 2 \Omega R \mathbf{\hat{x}}$

Kinetic energy. In the expression of the kinetic, you need to deal with the velocity of the material points. If the system has a point mass in $P$, its contribution to kinetic energy simply is

$$K_P = \frac{1}{2} m_P |\mathbf{v}_P|^2 = \frac{1}{2} m_P R^2 \Omega^2 \ .$$

If the system has a point mass in $D$ constrained on the surface of the disc, its contribution to kinetic energy is

$$K_D = \frac{1}{2} m_D |\mathbf{v}_D|^2 \ ,$$

where the velocity $\mathbf{v}_D$ is the time derivative of the position of $D$.

You need to discern material points and geometric points:

• let's call $T$ the geometric point identifying the top of the disc at all the times;
• let' call $P$ a material point, s.t. $P \equiv T$
• let's call $D$ a material point on the surface of the disc.

Here some facts will show below:

• The geometric point $T$ (and the material point $P \equiv T$) always has velocity $\mathbf{v}_T = \mathbf{v}_P = \Omega R \mathbf{\hat{x}}$;
• the material point $D$ has velocity $\mathbf{v}_D = 2 \Omega R \mathbf{\hat{x}}$ only when it passes through $T$;
• all the material points, when they pass through $T$, always has velocity $\mathbf{v} = 2 \Omega R \mathbf{\hat{x}}$

Kinetic energy. In the expression of the kinetic, you need to deal with the velocity of the material points. If the system has a point mass in $P$, its contribution to kinetic energy simply is

$$K_P = \frac{1}{2} m_P |\mathbf{v}_P|^2 = \frac{1}{2} m_P R^2 \Omega^2 \ .$$

If the system has a point mass in $D$ constrained on the surface of the disc, its contribution to kinetic energy is

$$K_D = \frac{1}{2} m_D |\mathbf{v}_D|^2 \ ,$$

where the velocity $\mathbf{v}_D$ is the time derivative of the position of $D$.

Kinematics. Velocity of points could be readily obtained as the time derivative of their position, $$\mathbf{v} = \dot{\mathbf{r}} \ .$$

Using Cartesian coordinates and the angle rotation angle $\theta$, s.t. $\Omega := \dot{\theta}$, and the assumption of pure rolling, the position and the velocity of points $P$, $D$ read:

$$\begin{aligned} & \begin{cases} \mathbf{r}_P = R \theta \mathbf{\hat{x}} + 2R \mathbf{\hat{y}} \\ \mathbf{r}_D = R \left(\theta - \sin \theta \right) \mathbf{\hat{x}}+ R \left(1-\cos \theta\right) \mathbf{\hat{y}} \\ \end{cases} \\ & \qquad \rightarrow \qquad \begin{cases} \mathbf{v}_P = R \Omega \mathbf{\hat{x}} \\ \mathbf{v}_D = R \Omega \left( 1 - \cos \theta \right) \mathbf{\hat{x}} + R \Omega \sin \theta \mathbf{\hat{y}}\\ \end{cases}\end{aligned} $$

Remark. It's easy to check that when $P \equiv D$, i.e. $\theta = \pi$, the velocity of $D$ reads

$$\mathbf{v}_P(\theta = \pi) = 2 R \Omega \mathbf{\hat{x}}$$