I have rewritten my answer to address the concerns made in comments by @juancarlosvegaoliver.
Instead of it being a rotation let it be simple harmonic motion in one dimension along the $x$-axis (vales increasing from left to right) and then I will show its relevance to the question asked.
The displacement from $O$ is $\vec x = x \hat x$ where $x$ is the component of the displacement in the direction of $\hat x$.
The velocity is $\dot x \hat x$ and the acceleration is $\ddot x \hat x$.
The force is $-k\vec x = -kx \hat x$ and $-kx$ is the component of the force in the direction of $\hat x$.
At a position like $A$ the displacement $\vec x$ in the direction of $\hat x$ and the direction of the (restoring) force is in the direction of $-\hat x$.
At position $B$ the displacement is in the direction of $-\hat x$ and the direction of the (restoring) force is in the direction of $+\hat x$.
So using $\vec F = m\vec a \Rightarrow -kx\hat x = m\ddot x \hat x \Rightarrow \ddot x = -\frac km x$ for all values of $x$ whether positive or negative.
Switching the direction of the unit vector to $\hat X = - \hat x$ makes no difference because now $\vec x = -x \hat X = -x \,(-\hat x) = x \hat x$ and $\vec a = -\ddot x \hat X = -\ddot x \,(-\hat x) = \ddot x \hat x$.
$\vec F = +kx\hat X = +kx (-\hat x) = -kx \hat x$.
For example, $2 \hat X = -2 \hat x$ and both give the position as $x=-2$
Having the $x$ axis pointing from right to left only results in a change of sign so, for example a position of $x=-2$ would now be $x=+2$.
Diagram 1 is the same as my diagram except that now $\vec x = x\hat x$ etc is replaced by $\theta \hat k$ etc and $\vec F - -k x \hat x$ is replaced by $\vec \tau = - mgd \sin \theta \,\hat k$.
The direction of $x$ increasing is to the right is replaced by the direction of $\theta$ increasing is anticlockwise.
Diagram 2 is just a reversal of the unit vector such that the new unit vector $\hat K = - \hat k$ with the direction of increasing $\theta$ still anticlockwise.
I have shown for the one dimension motion reversing the direction of the unit vector does not change anything and it is the same for the example with rotation.
$\vec \theta = -\theta \hat K = -\theta \,(-\hat k) = \theta \hat k$ and $\vec \alpha = -\ddot \theta \hat K = -\ddot \theta \,(-\hat k) = \ddot \theta \hat k$.
$\vec \tau = +mgd\hat K = +mgd (-\hat k) = -mgd \hat k$.