= Solution
For a <prime> $p$, the <divisor sum> defining the real weight is $\phi(0)-\phi(\log p/\log X)$. If $p>X^{1/3}$, the second term vanishes and the first is one. Thus \b[$F(p)=1$], while $F(n)\geq0$ for every integer because $\phi$ is real and the sum is squared.
Put $h(x)=e^x\phi(x)$. With the <Fourier transform> convention in the question, <Fourier inversion> gives
$$
\psi(t)=\frac1{2\pi}\int_{\mathbb R}h(x)e^{ixt}\,dx.
$$
The function $h$ is <smooth> and has <compact support>. Using <integration by parts> $k$ times, with zero endpoint terms, proves
$$
|\psi(t)|\leq\frac{\|h^{(k)}\|_1}{2\pi|t|^k}.
$$
Choose an integer $k\geq A$ to obtain \b[$|\psi(t)|\ll_A|t|^{-A}$ for $|t|\geq1$]. In particular every polynomially weighted absolute integral of $\psi$ is finite.
For the double integral, use
$$
\frac1{2+i(t+t')}=\int_0^\infty e^{-(2+i(t+t'))u}\,du,
\qquad
\phi(u)=\int_{\mathbb R}\psi(t)e^{-(1+it)u}\,dt.
$$
Differentiating the second formula gives $\int\psi(t)(1+it)e^{-(1+it)u}\,dt=-\phi'(u)$. <absolute convergence> from the rapid decay permits <Fubini's theorem>. The <derivative energy constant for a smooth sieve cutoff> is therefore
$$
\boxed{\iint\psi(t)\psi(t')\frac{(1+it)(1+it')}{2+i(t+t')}\,dt\,dt'
=c_\phi:=\int_0^\infty\phi'(u)^2\,du
=\int_0^{1/3}\phi'(u)^2\,du.}
$$
There is no <complex conjugate> in this integral: the two factors both become $-\phi'(u)$, which is real. The constant is positive; in fact the <Cauchy-Schwarz inequality> applied to $\phi(1/3)-\phi(0)=-1$ gives $c_\phi\geq3$.
Here is the basic structure of the <smooth divisor-square sieve asymptotic>, with enough uniform estimates to specify the error. Write $L=\log X$ and $R=X^{1/3}$. Expand the square and count multiples of the <least common multiple> $[d,e]$ in an arbitrary interval $I$ of length $X$:
$$
\sum_{n\in I}F(n)=X S_X+O_\phi(R^2),
\qquad
S_X=\sum_{d,e\geq1}\frac{\mu(d)\mu(e)}{[d,e]}\phi\left(\frac{\log d}L\right)\phi\left(\frac{\log e}L\right).
$$
The sum is effectively restricted to $d,e<R$. Each count is $X/[d,e]+O(1)$, independent of the interval's position, so the total error is $O_\phi(X^{2/3})$.
The <Fourier representation of a smooth Selberg weight> reads
$$
\phi\left(\frac{\log d}L\right)=\int_{\mathbb R}\psi(t)d^{-z}\,dt,
\qquad z=\frac{1+it}L.
$$
Hence, with $z'=(1+it')/L$,
$$
S_X=\iint\psi(t)\psi(t')E(z,z')\,dt\,dt',
\qquad
E(z,z')=\prod_p(1-p^{-1-z}-p^{-1-z'}+p^{-1-z-z'}).
$$
The four local terms correspond to the <prime> dividing neither <divisor>, only the first, only the second, or both. This is the <Euler product for a smoothed divisor-square correlation>. Factor it as
$$
E(z,z')=\frac{\zeta(1+z+z')}{\zeta(1+z)\zeta(1+z')}H(z,z'),
$$
where
$$
H(z,z')=\prod_p\frac{(1-a_p-b_p+c_p)(1-c_p)}{(1-a_p)(1-b_p)},
\quad a_p=p^{-1-z},\ b_p=p^{-1-z'},\ c_p=p^{-1-z-z'}.
$$
The local factors of $H$ are $1+O(p^{-2+4\eta})$ when $|\operatorname{Re}z|,|\operatorname{Re}z'|\leq\eta$ for small fixed $\eta<1/4$. Thus $H$ is a <holomorphic function> near $(0,0)$, and evaluating each local factor at zero gives exactly one. Consequently $H(0,0)=1$ and $H(z,z')=1+O(|z|+|z'|)$ there.
The pole expansion $\zeta(1+w)=w^{-1}+O(1)$ now yields
$$
E(z,z')=\frac1L\frac{(1+it)(1+it')}{2+i(t+t')}
\left(1+O\left(\frac{2+|t|+|t'|}L\right)\right)
$$
when $|t|,|t'|\leq L^{1/4}$. Integrating the error against the rapidly decreasing transforms gives $O_\phi(L^{-2})$. The complementary tails are harmless uniformly: absolute values in the original <divisor> series give
$$
|E((1+it)/L,(1+it')/L)|
\leq\prod_p(1+2p^{-1-1/L}+p^{-1-2/L})
\leq\zeta(1+1/L)^3\ll L^3.
$$
Any prescribed power decay of $\int_{|t|>L^{1/4}}|\psi(t)|\,dt$ is available, so these tails and the tails of the limiting kernel are $O_\phi(L^{-2})$ after choosing sufficiently many applications of <integration by parts>. The same absolute bound justifies the original exchanges of infinite sums and integrals. Using the evaluated double integral,
$$
S_X=\frac{c_\phi}L+O_\phi(L^{-2}).
$$
We conclude, uniformly over all interval locations,
$$
\boxed{\sum_{n\in I}F(n)=c_\phi\frac X{\log X}
+O_\phi\left(\frac X{\log^2X}+X^{2/3}\right)
\sim c_\phi\frac X{\log X}.}
$$
This proof does not assume that the interval starts near the origin.
Finally choose this one fixed <smooth> cutoff. <primes> above $R$ in $(X_0,X_0+X]$ contribute one each to the nonnegative weight, and at most $R$ <primes> lie below $R$. Therefore
$$
\pi(X_0+X)-\pi(X_0)\leq\sum_{X_0<n\leq X_0+X}F(n)+R
\ll\frac X{\log X}
$$
for all sufficiently large $X$, uniformly in $X_0$. The bounded range $2\leq X\leq X_1$ follows by increasing the fixed constant and using the trivial interval bound $X+1$. Thus the <short-interval prime upper bound from a smooth divisor weight> is
$$
\boxed{\pi(X_0+X)-\pi(X_0)\ll\frac X{\log X}\qquad(X_0,X\geq2).}
$$
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