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.  

[![enter image description here][1]][1]

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 $\vec \alpha = -\ddot \theta  \hat K = -\ddot \theta \,(-\hat k) = \ddot \theta \hat k$.  
$\vec \tau = +mgd\hat K = +mgd (-\hat k) = -mgd \hat k$.  


  [1]: https://i.sstatic.net/dzvx5.jpg