Surely I'm making a trivial calculation error but I can not find it.
Let: $$L(r,\theta,\phi)= \frac{\hbar}{i} r \left( \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} \hat{\phi} - \frac{1}{r}\frac{\partial }{\partial \theta} \hat{\theta} \right)$$ $$L^2=L\cdot L= -\hbar^2 r^2 \left( \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} \hat{\phi} - \frac{1}{r}\frac{\partial }{\partial \theta} \hat{\theta} \right)^2=$$ $$= -\hbar^2 r^2 \left( \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} \hat{\phi} - \frac{1}{r}\frac{\partial }{\partial \theta} \hat{\theta} \right) \cdot \left( \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} \hat{\phi} - \frac{1}{r}\frac{\partial }{\partial \theta} \hat{\theta} \right)=$$ $$-\hbar^2 \left( \frac{1}{\sin^2(\theta)} \frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial \theta^2} \right)$$
But I know that the correct result is: $$L^2=-\hbar^2 \left( \frac{1}{\sin^2(\theta)} \frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial \theta^2} + \cot\theta \frac{\partial}{\partial \theta} \right)$$
The problem is that if you change $\theta$, the direction of $\hat\phi$ changes. You should be able to prove that $$\frac{\partial \hat\theta}{\partial \phi}=\hat\phi\cos\theta$$ If you plug this into your equation, you will get the missing term.