What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2?

The equation you wrote doesn't mention anything about the linear acceleration.

Is the center of mass located at its top and not the center? Or is there something else I am missing?

The center of mass of the pencil is in the middle, not the top. There is likely something else you are missing. Or, rather, maybe there is something else the person who wrote what you are reading is missing.

The tipping pencil has many forces acting on it: the force of gravity acting at the center of mass; the normal force acting at the bottom; the frictional force acting at the bottom.

As far as I can tell, the easiest way to solve this problem is by applying a sum of torques analysis with the axis of rotation chosen as the bottom of the pencil. Then the sum of torques gives (the gravitation force acts at the center of mass): $$ \frac{mg l \sin(\theta)}{2} $$ and this is equal to $$ I\frac{d^2\theta}{dt^2} $$ where $I$ is the moment of inertia of a pencil about it bottom (not its middle, because the bottom was our choice for the axis of rotation) $$ I=\frac{ml^2}{3} $$ which, for small angles, gives $$ \frac{d^2\theta}{dt^2}=\frac{3}{2}\frac{g \theta}{l} $$ which is different from your equation by a factor of 3/2.

If you consider the pencil not to be a solid rod, but rather to be a point mass located at the center of mass (at l/2) then you get $I=ml^2/2$ and you recover the equation you had written.

What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2?

The equation you wrote doesn't mention anything about the linear acceleration.

Is the center of mass located at its top and not the center? Or is there something else I am missing?

The center of mass of the pencil is in the middle, not the top. There is likely something else you are missing. Or, rather, maybe there is something else the person who wrote what you are reading is missing.

The tipping pencil has many forces acting on it: the force of gravity acting at the center of mass; the normal force acting at the bottom; the frictional force acting at the bottom.

As far as I can tell, the easiest way to solve this problem is by applying a sum of torques analysis with the axis of rotation chosen as the bottom of the pencil. Then the sum of torques gives (the gravitation force acts at the center of mass): $$ \frac{mg l \sin(\theta)}{2} $$ and this is equal to $$ I\frac{d^2\theta}{dt^2} $$ where $I$ is the moment of inertia of a pencil about it bottom (not its middle, because the bottom was our choice for the axis of rotation) $$ I=\frac{ml^2}{3} $$ which, for small angles, gives $$ \frac{d^2\theta}{dt^2}=\frac{3}{2}\frac{g \theta}{l} $$ which is different from your equation by a factor of 3/2.

If you consider the pencil not to be a solid rod, but rather to be a point mass located at the center of mass (at l/2) then you get $I=ml^2/4$ and you recover the equation you had written.

What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2?

The equation you wrote doesn't mention anything about the linear acceleration.

Is the center of mass located at its top and not the center? Or is there something else I am missing?

The center of mass of the pencil is in the middle, not the top. There is likely something else you are missing. Or, rather, maybe there is something else the person who wrote what you are reading is missing.

The tipping pencil has many forces acting on it: the force of gravity acting at the center of mass; the normal force acting at the bottom; the frictional force acting at the bottom.

As far as I can tell, the easiest way to solve this problem is by applying a sum of torques analysis with the axis of rotation chosen as the bottom of the pencil. Then the sum of torques gives (the gravitation force acts at the center of mass): $$ \frac{mg l \sin(\theta)}{2} $$ and this is equal to $$ I\frac{d^2\theta}{dt^2} $$ where $I$ is the moment of inertia of a pencil about it bottom (not its middle, because the bottom was our choice for the axis of rotation) $$ I=\frac{ml^2}{3} $$ which, for small angles, gives $$ \frac{d^2\theta}{dt^2}=\frac{3}{2}\frac{g \theta}{l} $$ which is different from your equation by a factor of 3/2.

What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2?

The equation you wrote doesn't mention anything about the linear acceleration.

Is the center of mass located at its top and not the center? Or is there something else I am missing?

The center of mass of the pencil is in the middle, not the top. There is likely something else you are missing. Or, rather, maybe there is something else the person who wrote what you are reading is missing.

The tipping pencil has many forces acting on it: the force of gravity acting at the center of mass; the normal force acting at the bottom; the frictional force acting at the bottom.

As far as I can tell, the easiest way to solve this problem is by applying a sum of torques analysis with the axis of rotation chosen as the bottom of the pencil. Then the sum of torques gives (the gravitation force acts at the center of mass): $$ \frac{mg l \sin(\theta)}{2} $$ and this is equal to $$ I\frac{d^2\theta}{dt^2} $$ where $I$ is the moment of inertia of a pencil about it bottom (not its middle, because the bottom was our choice for the axis of rotation) $$ I=\frac{ml^2}{3} $$ which, for small angles, gives $$ \frac{d^2\theta}{dt^2}=\frac{3}{2}\frac{g \theta}{l} $$ which is different from your equation by a factor of 3/2.

If you consider the pencil not to be a solid rod, but rather to be a point mass located at the center of mass (at l/2) then you get $I=ml^2/2$ and you recover the equation you had written.