Suppose $\Phi$ is a function of $r, \theta$ and $\phi$. If I want to derive the Laplacian for this function, I would assume that..
$$\nabla ^2 \Phi = \nabla \cdot \nabla \Phi$$
And as, in spherical:
$$\nabla = \frac{\partial}{\partial r} \hat r + \frac{1}{r} \frac{\partial}{\partial \theta} \hat \theta + \frac{1}{r \sin \ \theta} \frac{\partial}{\partial \phi} \hat \phi$$
Which implies
$$\nabla \Phi= \frac{\partial \Phi}{\partial r} \hat r + \frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat \theta + \frac{1}{r \sin \ \theta} \frac{\partial \Phi}{\partial \phi} \hat \phi$$
Thus, the Laplacian would seem to me to be:
$$\large\begin{bmatrix} \frac{\partial}{\partial r} \\ \frac{1}{r} \frac{\partial}{\partial \theta} \\ \frac{1}{r \sin \ \theta} \frac{\partial}{\partial \phi} \\ \end{bmatrix} \cdot \begin{bmatrix} \frac{\partial \Phi}{\partial r} \\ \frac{1}{r} \frac{\partial \Phi}{\partial \theta} \\ \frac{1}{r \sin \ \theta} \frac{\partial \Phi}{\partial \phi} \\ \end{bmatrix}$$
Which will not give me the form for the Laplacian in spherical coordinates in my lecture notes or the internet. Where do I have it wrong?
The Laplacian is the divergence of the gradient. And the divergence in spherical coordinates is: $$\nabla\cdot \mathbf A = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( A_\theta\sin\theta \right) + {1 \over r\sin\theta}{\partial A_\varphi \over \partial \varphi}$$
Now substitute the $\nabla\Phi$ that you already have for $\mathbf A$.
That leaves the question how we got that formula for the divergence. You can find the derivation here: Nabla in spherical coordinates. It explains it in terms of how divergence is defined.