I don't really understand the standard result in Holography and AdS/CFT that a source field can be expressed at \begin{equation} \phi_0 = \lim_{z \rightarrow 0} z^{\Delta -d} \phi(z,x) \end{equation} because I don't understand some steps in the derivation. $\Delta$ is the dimension of the corresponding field theory operator $O$. $d$ is the spatial dimension of the AdS spacetime. I have a number of beginner questions when deriving the above.
Usually, by e.g. following these notes: https://arxiv.org/abs/1908.02667, we consider the background metric
\begin{equation} ds^2 = \frac{L^2}{z^2} \Big( dz^2 - dt^2 -dx^2 \Big). \end{equation}
First of all, what does it mean that the AdS boundary is located at $z \to 0$? I don't see any intuition behind this at this stage?
From the action for a massive scalar
\begin{equation} S= -\frac{1}{2} \int d^{d+1}x \sqrt{-g} \Big[ g^{MN} \partial_M \phi \partial_N \phi + m^2 \phi^2 \Big]. \end{equation} By varying this action we get the equation of motion \begin{equation} \frac{1}{\sqrt{-g}} \partial_M \Big( \sqrt{-g}g^{MN} \partial_M \phi \Big) - m^2\phi^2 = 0. \end{equation} Now, my second question is what it means to write this explicitly on the geometric background defined by the metric above? The above notes state that this yields \begin{equation} z^{d+1} \partial_z \Big( z^{1-d} \partial_z \phi \Big) + z^2 \delta^{\mu \nu} \partial_\mu \partial_\nu \phi - m^2 L^2 \phi =0. \end{equation} Maybe it is obvious but I don't see where this equation comes from or what it means? Would love to see this explicitly and understand it.