I) L&L are referring to a linearized model where the TISE
$$\begin{align} \hbar^2\psi^{\prime\prime}(x) +P^2(x) \psi(x)~=~&0,\cr P(x)~:=~ \sqrt{2m (E-V(x))}~=~& |P(x)| e^{i\theta(x)}, \end{align} \tag{1}$$
becomes the Airy differential equation. Near a turning point $x_0$, we can approximate the square momentum
$$\begin{align} P^2(x)~\approx~& \alpha (x\!-\!x_0), \cr (P^2)^{\prime}(x)~\approx~&\alpha~=~2mF_0 ~\neq~0,\end{align} \tag{2} $$
with an affine function, where $\alpha\in \mathbb{R}\backslash\{0\}$ is a non-zero real constant. (In L&L's example $\alpha < 0$ is negative).
II) Let us first reproduce L&L's argument. In the classically allowed region $x<x_0$, L&L choose the momentum $P^{>}>0$ positive. (The superscripts $>$ and $<$ refer to the classically allowed and forbidden regions, respectively.) Hence the WKB wave function (47.2) becomes
$$\begin{align} \psi^{>}~\stackrel{(47.2)}{=}&~\frac{1}{\sqrt{P^{>}}}\sum_{\pm}C_{\pm} \exp\left(\pm \frac{i}{\hbar}\int_{x_0}^x \! P^{>}~\mathrm{d}x\right) \cr
~=~&\frac{1}{\sqrt{|P|}} \sum_{\pm}C_{\pm} \exp\left( \mp \frac{2i\sqrt{|\alpha|}}{3\hbar}\left(x_0-x \right)^{\frac{3}{2}}\right)\cr
& \qquad\text{for}\qquad x<x_0.\end{align}\tag{3}$$
III) L&L is next analytically continuing the WKB wave function (47.1) (which is valid in the classically forbidden region) along an (upper/lower) semicircle in the complex $x$-plane
$$ x-x_0~=~\rho e^{\pm i\phi} , \qquad \phi~\in~[0,\pi], \tag{47.4a}$$
in order to derive the WKB connection formulas.
L&L choose the momentum variable
$$\begin{align} P^{<}~=~&\sqrt{|\alpha|(x-x_0)}\cr
~=~&\sqrt{|\alpha|}\rho^{\frac{1}{2}} e^{\pm\frac{ i\phi}{2}}, \qquad \rho~>~0,\end{align}\tag{4} $$
such that it starts out positive
$$ \left. P^{<}\right|_{\phi=0}
~=~|P|~>~0. \tag{5} $$
The analytical continuation of the WKB wave function (47.1) then becomes
$$\begin{align} \psi^{<}~\stackrel{(47.1)}{=}&~\frac{C}{2\sqrt{P^{<}}} \exp\left( -\frac{1}{\hbar}\int_{x_0}^x \! P^{<}~\mathrm{d}x\right)\cr
~=~&\frac{C}{2\sqrt{|P|}} \exp\left( -\frac{2\sqrt{|\alpha|}\rho^{\frac{3}{2}}e^{\pm\frac{ i3\phi}{2}} }{3\hbar} \mp i\frac{\phi}{4}\right).
\end{align}\tag{6} $$
By setting $\phi=\pi$ in eq. (6), and comparing with the WKB wave function (3), one deduces from the analytical continuation along the (upper/lower) semicircle that
$$\begin{align} C_2~\equiv~& C_-~=~\frac{C}{2}e^{-\frac{i\pi}{4}}, \tag{47.4b}\cr
C_1~\equiv~& C_+~=~\frac{C}{2}e^{\frac{i\pi}{4}} , \tag{47.4c}\end{align} $$
respectively. This is L&L's main argument. (We will return to the question of what happens to the other branch in Section VII below.)
IV) Alternatively, since the Airy functions have Fourier integral representations (cf. e.g. my Math.SE answer here), one may instead use the method of steepest descent to derive matching asymptotic expansions on each side of the turning point. It is interesting to compare this method with L&L's above argument.
One may show that the Airy-like wave function
$$ \psi(x) ~=~ \sqrt{\frac{|\alpha|}{\pi\hbar}}\int_{\mathbb{R}-i~{\rm sgn}(\alpha)~\varepsilon} \! \mathrm{d}p~
\exp\left(\frac{i}{\hbar}S(p,x)\right),\tag{7}$$
$$\begin{align} S(p,x)~:=~& px - \tilde{S}(p), \cr
\tilde{S}(p)~:=~& p\left(x_0+\frac{p^2}{3\alpha}\right),\end{align} \tag{8}$$
$$\begin{align} S^{\prime}(p,x)~=~&x-\tilde{S}^{\prime}(p)\cr ~=~& x\!-\!x_0 - \frac{p^2}{\alpha}\cr ~\stackrel{(2)}{\approx}~& \frac{P^2(x)-p^2}{\alpha},\end{align} \tag{9} $$
$$-S^{\prime\prime}(p,x)~=~\tilde{S}^{\prime\prime}(p)~=~ \frac{2p}{\alpha},\tag{10} $$
satisfies the TISE (1):
$$\begin{align} \hbar^2& \psi^{\prime\prime}(x) +P^2(x) \psi(x)\cr
~\stackrel{(7)}{=}~&
\sqrt{\frac{|\alpha|}{\pi\hbar}}
\int_{\mathbb{R}-i~{\rm sgn}(\alpha)~\varepsilon} \! \mathrm{d}p~\left(P^2(x)- p^2\right) \exp\left(\frac{i}{\hbar}S(p,x)\right)\cr
~\stackrel{(9)}{\approx}~&
-i \hbar \alpha\sqrt{\frac{|\alpha|}{\pi\hbar}}
\int_{\mathbb{R}-i~{\rm sgn}(\alpha)~\varepsilon} \! \mathrm{d}p~\frac{d}{dp} \exp\left(\frac{i}{\hbar}S(p,x)\right)\cr
~=~&0 \end{align}\tag{11}$$
for the chosen contour in the complex $p$-plane.
V) There are 2 critical points:
$$\begin{align} p_{\sigma} ~=~& \sigma~ {\rm sgn}(\alpha)~P , \cr
S_{\sigma}~=~& \frac{2\sigma P^3 }{3|\alpha|} , \cr
S^{\prime\prime}_{\sigma}~=~& -2\sigma |\alpha| P , \cr
\sigma ~\in~&\{\pm 1\}.\end{align} \tag{12}$$
The steepest descent contribution from each critical point comes from a Gaussian integral:
$$\begin{align} I_{\sigma}~=~& \sqrt{\frac{|\alpha|}{\pi\hbar}} \sqrt{\frac{2\pi i\hbar}{S^{\prime\prime}_{\sigma} } } \exp\left(\frac{i S_{\sigma}}{\hbar}\right)\cr
~=~&\frac{1}{\sqrt{P}} \exp\left(\frac{2i\sigma P^3}{3\hbar |\alpha|} -\frac{i\sigma \pi}{4}\right)\cr
~=~& \frac{1}{\sqrt{|P|}} \exp\left(\frac{2i\sigma |P|^3e^{3i\theta}}{3\hbar |\alpha|} -i\left(\frac{\sigma \pi}{4}+\frac{\theta}{2}\right)\right) \end{align} \tag{13} $$
with angular steepest descent direction $-\frac{\sigma \pi}{4}-\frac{\theta}{2}$.
$\uparrow$ Fig. 1. The complex $p$-plane with 3 possible integration contours ${\cal C}_1$, ${\cal C}_2$, ${\cal C}_3$ for $\alpha<0$. The shaded regions denote exponentially decaying sectors. (If $\alpha>0$, it is opposite.) Since the integrand is an entire function in $p$, the $p$-integral only depend on the monodromy of the contour. (Figure taken from Ref. [W].)
VI) Now let us return to L&L's example with $\alpha<0$.
Classically forbidden region $x>x_0$: Then $P=\pm i|P|$, with $\theta=\pm \frac{\pi}{2}$, where $\pm$ corresponds to two different possible sign choices. The contour along the real $p$-axis is reproduced by the steepest descent contour corresponding to the critical point $p_{\mp 1}$:
$$\begin{align} \psi_{<}(x)~\sim~& I_{\mp 1}\cr
~=~& \frac{1}{\sqrt{|P|}} \exp\left(-\frac{2|P|^3}{3\hbar |\alpha|} \right). \end{align} \tag{14} $$
Classically allowed region $x<x_0$: Then $P=|P|>0$ with $\theta=0$. The contour along the real $p$-axis is reproduced by the sum of both steepest descent contours:
$$\begin{align} \psi_{>}(x)~\sim~& \sum_{\sigma\in\{\pm 1\}}I_{\sigma}\cr
~=~& \frac{1}{\sqrt{|P|}}\sum_{\sigma\in\{\pm 1\}} \exp\left(i\sigma\left(\frac{2|P|^3}{3\hbar |\alpha|} -\frac{ \pi}{4}\right)\right)\cr
~=~&\frac{2}{\sqrt{|P|}} \cos\left(\frac{2|P|^3}{3\hbar |\alpha|} -\frac{\pi}{4} \right)\cr
~=~& \frac{2}{\sqrt{|P|}} \sin\left(\frac{2|P|^3}{3\hbar |\alpha|} +\frac{\pi}{4} \right).\end{align} \tag{15} $$
By comparing eqs. (14) & (15), we have derived a connection formula. In detail, by comparing the contributions to the critical points $p_{\mp 1}$, we deduce L&L's eqs. (47.4c) and (47.4b), respectively. So the two methods agree.
VII) Exponentially growing solutions are physically possible if the classically forbidden region has finite length, e.g. in quantum tunneling. However, in L&L's example, the classically forbidden region is non-compact, so exponentially growing solutions must be discarded.
It remains to explain why only one of the 2 oscillatory branches in the classically allowed side gets analytically continued to the classically forbidden side.
On one hand, L&L argue that the exponentially growing WKB branch cannot be trusted during the $\frac{3\pi}{2}$ phase shift in the Boltzmann factor because it becomes momentarily exponentially suppressed. This highlights the fact that the connection formulas are one-directional.
On the other hand, from the perspective of the method of steepest descent in Sections IV-VI, one crosses a Stokes line in the complex $x$-plane, so that the steepest descent contours pick up monodromy in the complex $p$-plane, cf. Fig. 1.
For more information, see also e.g. my related Phys.SE answer here.
VIII) Finally, we should mention that the $\frac{\pi}{2}$ phase shift between the incoming and outgoing wave at the turning point in eq. (15) corresponds to a Maslov index $|\mu|=1$,
cf. e.g. this Phys.SE post.
References:
[LL] L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 2nd & 3rd ed, 1981; $\S47$ & $\S b$.
[W] E. Witten, Analytic Continuation Of Chern-Simons Theory, arXiv:1001.2933; p. 23-29, 48-49. A related 2015 KITP lecture by Witten, A New Look At The Path Integral Of Quantum Mechanics, can be found on YouTube.
