= Solution
Reflect $r'=(x',y',z')$ to $r'^*=(x',y',-z')$. The <method of images> gives the <Dirichlet image kernel for the three-dimensional Helmholtz equation>:
$$
\boxed{G_D(r,r')=-\frac1{4\pi}\left[\frac{e^{ik|r-r'|}}{|r-r'|}-\frac{e^{ik|r-r'^*|}}{|r-r'^*|}\right].}
$$
The two distances coincide on the plane, so its boundary value is zero; the image source is outside the domain. It therefore has the required positive delta source inside and the selected outgoing behaviour.
Let $h(R)=e^{ikR}/R$. At a boundary point $r=(\xi,\eta,0)$, the distances to the real and reflected sources have $z$ derivatives $-z'/R$ and $+z'/R$. Thus
$$
\partial_zG_D=\frac{z'}{2\pi R}h'(R),\qquad h'(R)=\frac{e^{ikR}}R\left(ik-\frac1R\right).
$$
The outward <unit normal> of the upper half-space is $-e_z$, and $z'/R=\cos\theta$. Hence
$$
\partial_nG_D=-\frac{ik}{2\pi}\frac{e^{ikR}}R\left(1+\frac{i}{kR}\right)\cos\theta.
$$
Using the decay assumption to discard the distant hemisphere in the boundary representation gives
$$
\boxed{\phi(r')=-\frac{ik}{2\pi}\int_{z=0}\frac{e^{ikR}}R\left(1+\frac{i}{kR}\right)\cos\theta\,\phi(r)\,dS.}
$$
The qualification at infinity concerns the whole Green boundary term, not just the pointwise limit of $\phi$. Sufficient decay of the field and its derivatives, as assumed, makes it vanish; compatible radiation conditions can also do so. As a sign check, taking $k\to0$ in the kernel gives $z'/(2\pi R^3)$, the positive <Poisson kernel for the upper half-space>.
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