Derivation of the basic equation for Witten diagrams I could understand the derivation of the "bulk-to-boundary" propagators ($K$) for scalar fields in $AdS$ but the iterative definition of the "bulk-to-bulk" propagators is not clear to me. 
On is using the notation that $K^{\Delta_i}(z,x;x')$ is the bulk-to-boundary propagator i.e it solves $(\Box -m^2)K^{\Delta_i}(z,x;x') = \delta (x-x')$ and it decays as $cz^{-\Delta _i}$ (for some constant $c$) for $z \rightarrow 0$. Specifically one has the expression, $K^{\Delta_i}(z,x;x') =  c \frac {z^{\Delta _i}}{(z^2 + (x-x')^2)^{\Delta_i}}$


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*Given that this $K$ is integrated with boundary fields at $x'$ to get a bulk field at $(z,x)$, I don't understand why this is called a bulk-to-boundary propagator. I would have thought that this is the "boundary-to-bulk" propagator! I would be glad if someone can explain this terminology. 

*Though the following equation is very intuitive, I am unable to find a derivation for this and I want to know the derivation for this more generalized expression which is written as,
$\phi_i(z,x) = \int d^Dx'K^{\Delta_i}(z,x;x')\phi^0_i(x') + b\int d^Dx' dz' \sqrt{-g}G^{\Delta_i}(z,x;z',x') \times$
$\int d^D x_1 \int d^D x_2 K^{\Delta_j}(z,x;x_1)K^{\Delta_k}(z,x;x_2)\phi^0_j(x_1) \phi^)_k(x_2) + ...$
where the "b" is as defined below in the action $S_{bulk}$, the fields with superscript of $^0$ are possibly the values of the fields at the boundary and  $G^{\Delta_i}(z,x;z',x')$ - the "bulk-to-bulk" propagator is defined as the function such that,
$(\Box - m_i^2)G^{\Delta_i}(z,x;z',x') = \frac{1}{\sqrt{-g}} \delta(z-z')\delta^D(x-x')$


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*Here what is the limiting value of this $G^{\Delta_i}(z,x;z',x')$ that justifies the subscript of $\Delta_i$. 


Also in this context one redefined $K(z,x;x')$ as,
$K(z,x;x') = lim _ {z' \rightarrow 0} \frac{1}{\sqrt{\gamma}} \vec{n}.\partial G(z,x;z',x')$
where $\gamma$ is the metric $g$ restricted to the boundary. 


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*How does one show that this definition of $K$ and the one given before are the same? (..though its very intuitive..)   

*I would also like to know if the above generalized expression is somehow tied to the following specific form of the Lagrangian, 
$S_{bulk} = \frac{1}{2} \int d^{D+1}x \sqrt{-g} \left [ \sum _{i=1}^3 \left\{ (\partial \phi)^2 + m^2 \phi_i^2 \right\} + b \phi_1\phi_2 \phi_3 \right ]$ 
Is it necessary that for the above expression to be true one needs multiple fields/species? Isn't the equation below the italicized question a general expression for any scalar field theory in any space-time? 


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*Is there a general way to derive such propagator equations for lagrangians of fields which keep track of the behaviour at the boundary? 

 A: Your first question is just semantics; I agree that boundary-to-bulk is more intuitive.
The equation below your italicized question is the iterative solution to the field equations of the $\phi^3$ Lagrangian, to first order in the coupling constant. It would in general depend on the specific bulk theory. For example, if you had a $\lambda \phi^4$ interaction, you would find
\begin{align*}
\phi=\int dx  K\phi^0+\lambda\int dz dx G\int dx_1dx_2dx_3 K_1K_2K_3\phi^0_1\phi_2^0\phi_3^0+\ldots,
\end{align*}
and so on. 
The reason that finding the classical solution to the field equations is useful is that in the semiclassical limit, the bulk path integral with the source $\phi_0$ turned on can be written as
\begin{align*}
Z[\phi_0]=\int d\phi|_{\phi(z=\epsilon)=\epsilon^{d-\Delta}\phi_0} \exp(-S[\phi])\sim \exp(-S[\phi_{\text{cl}}]),
\end{align*}
where $\phi_{\text{cl}}$ is the extension of $\phi_0$ to a solution of the bulk field equations which is regular at the horizon. Differentiating both sides with respect to $\phi_0$ then evaluates the tree level Feynman diagrams of the boundary theory.
I think that the notation $G^{\Delta_i}$ is meant to show that $G$ depends on the masses of the fields (and therefore the dual dimensions of the bulk operators). Its limiting behavior can be seen from the next equation that you wrote, which in turn follows from Stokes' theorem.
