Assume first only that $f > 0$ and that $\log(f)$ is concave. Define the function
\begin{equation}
I(x,y,u) =
\begin{cases}
1 &\quad\text{if} & x > u > y\\
0 &\quad\text{otherwise.} \\
\end{cases}
\end{equation}
Since the set $\{(x,y,u) : I(x,y,u) = 1 \}$ is convex, $I$ is log-concave and so also the function $(x,y,u) \in \mathbb{R}^3 \mapsto h(x,y,u)f(u)$ is log-concave. By the Theorem quoted in the post we have then that
\begin{equation}
H(x,y)= \int_{\mathbb{R}} I(x,y,u)f(u) du
\end{equation}
is log-concave. Since for $x > y$, we have $H(x,y)=F(x) - F(y) > 0$, we get that the function $G(x,y)= \log(H(x,y))$, defined for $x > y$, is concave. This is Pratt's proof quoted in Note (2).
Now, let us consider the hypotheses in the post, that is $f > 0$ and $\log(f)$ strictly concave. We must show that $G$ turns out to be strictly concave. This follows by applying to our particular case the argument given in Brascamp and Lieb, Some inequalities for Gaussian measures and the long-range order of the one-dimensional plasma to prove the Theorem stated in the post.
We choose $x > y$ and $x'> y'$, with $(x,y) \neq (x',y')$, and we consider the log-concave function $Q: \mathbb{R}^2 \rightarrow [0,\infty)$ defined by
\begin{equation}
Q(t,u)=I(t x + (1-t)x',t y + (1-t)y',u)f(u),
\end{equation}
so that
$H(t x + (1-t)x',t y + (1-t)y')=P(t)= \int_{\mathbb{R}} Q(t,u) du$.
Consider $t=1, t'=0$, fix $\lambda \in (0,1)$, and let $b \in \mathbb{R}$ such that $\sup_{u \in \mathbb{R}} Q(0,u)= e^{b} \sup_{u \in \mathbb{R}} Q(1,u)$. Define the log-concave function $\bar{Q}(t,u)=e^{bt}Q(t,u)$, and put
\begin{equation}
\bar{P}(t)=\int_{\mathbb{R}}\bar{Q}(t,u)du=e^{bt}P(t).
\end{equation}
Clearly, $P(t)$ is strictly log-concave if and only if $\bar{P}(t)$ is so.
Now, let us put, for any fixed $z \geq 0$:
\begin{equation}
C(z)= \{ (t,u) \in [0,1] \times \mathbb{R} : \bar{Q}(t,u) \geq z \},
\end{equation}
and for any fixed $(t,z) \in [0,1] \times [0,\infty)$:
\begin{equation}
C(t,z)= \{ t \} \times \{u \in \mathbb{R}: \bar{Q}(t,u) \geq z \}.
\end{equation}
$C(z)$ is a convex set, and $C(t,z)$ is a segment.
Then let $T(z)$ be the trapezoid having as parallel sides the segments $C(0,z)$ and $C(1,z)$.
$C(z)$ is the intersection of $T(0)$ with the set
\begin{equation}
B(z) = \{ (t,u) \in [0,1] \times \mathbb{R} : e^{bt} f(u) \geq z \},
\end{equation}
which is convex since $(t,u) \mapsto e^{bt}f(u)$ is (strictly) log-concave.
Now, let us note that since $\log(f)$ is strictly log-concave, in particular it is not constant on any interval, so that $C(0,z)$ and
$C(1,z)$ shrink simultaneously to a single point when $z$ approaches $\bar{z}=\sup_{u \in \mathbb{R}}Q(0,u)$. We have then, by convexity of $C(z)$, that if $g(t,z)$ is the length of the segment $C(t,z)$, we have for any $z \geq 0$:
\begin{equation}
g(\lambda,z) \geq \lambda g(0,z) + (1- \lambda) g(1,z) \quad (I).
\end{equation}
From a well known result about distribution functions (see e.g. Rudin, Real and Complex Analysis, 3rd Edition, Section (18.5)) if $\mu$ is the one dimensional Lebesgue measure, we have
\begin{equation}
\bar{P}(t)=\int_{\mathbb{R}} \bar{Q}(t,u)du=\int_{0}^{\infty} \mu(\{u : \bar{Q}(t,u) > z \}) dz = \int_{0}^{\infty} \mu(\{u : \bar{Q}(t,u) \geq z \}) dz = \int_{0}^{\infty} g(t,z) dz,
\end{equation}
(where the third equality follows by continuity of $f$ or more in general because the nonincreasing functions $z \mapsto \mu(\{u : \bar{Q}(t,u) > z \})$ and $z \mapsto \mu(\{u : \bar{Q}(t,u) \geq z \})$ differ only on a countable set).
We then get
\begin{equation}
\bar{P}(\lambda) \geq \lambda \bar{P}(0) + (1 - \lambda) \bar{P}(1) \quad (II).
\end{equation}
Now assume that (II) holds with equality. Set for any $t \in [0,1]$:
\begin{equation}
L(t,z)= \inf \{u \in \mathbb{R} : \bar{Q}(t,u) \geq z \},
\end{equation}
\begin{equation}
L(z)=\{(t,L(t,z)) \in \mathbb{R}^2 : t \in [0,1] \},
\end{equation}
\begin{equation}
U(t,z)= \sup \{u \in \mathbb{R}: \bar{Q}(t,u) \geq z \},
\end{equation}
\begin{equation}
U(z)=\{(t,U(t,z)) \in \mathbb{R}^2 : t \in [0,1] \},
\end{equation}
and note that $L(0,z)$, $L(1,z)$, $U(0,z)$ and $U(1,z)$ are the vertices of $T(z)$ (in particular $L(0,0)=y$, $L(1,0)=y'$, $U(0,0)=x$ and $U(1,0)=x'$) and that $z \mapsto L(t,z)$ and $z \mapsto U(t,z)$ are continuous.
Since (II) holds with equality, (I) must hold with equality for a.e. $z \in [0,\bar{z}]$, and so actually for every $z \in [0,\bar{z}]$, given the contintuity of $z \mapsto g(t,z)$. This in turn implies that for every $z \in [0,\bar{z}]$ $L(z)$ and $U(z)$ must be segments, so we conlude that $C(z)$ must coincide with the trapezoid $T(z)$ for all $z \in [0,\bar{z}]$.
But then, since for $z$ approaching $\bar{z}$ from the left, $C(0,z)$ and $C(1,z)$ shrink to single points and since $(x,y) \neq (x',y')$, there exists some $z \in (0,\bar{z}]$ such that at least one between the segments $L(z)$ and $U(z)$ is not horizontal and not coinciding with a side of $T(0)$. Fix such a $z$ and let e.g. $L(z)$ be the segment with the stated property. Then we have
\begin{equation}
z=e^{bt}f(u)
\end{equation}
for every $(t,u) \in L(z)$, so that if $t=c+mu$ is the equation of the straight line containing $L(z)$, we get
\begin{equation}
f(u)= z \exp(-bc-bmu),
\end{equation}
for $u$ ranging between $-c/m$ and $(1-c)/m$. So $f$ would be not strictly log-concave, a contradiction.
We conclude that (II) holds strictly, and so
\begin{equation}
\bar{P}(\lambda) > \lambda \bar{P}(0) + (1 - \lambda) \bar{P}(1) \geq [\bar{P}(0)]^{\lambda} [\bar{P}(1)]^{1-\lambda}
\end{equation}
where the last inequality is the classical inequality between arithmetic mean and geometric mean.
So we have
\begin{equation}
P(\lambda) > [P(0)]^{\lambda} [P(1)]^{1-\lambda},
\end{equation}
that is, by taking the logarithms,
\begin{equation}
G(\lambda x + (1 - \lambda) x', \lambda y + (1- \lambda) y') > \lambda G(x,y) + (1 - \lambda) G(x',y').
\end{equation}
QED
