= Solution
Use an incoming spacelike <momentum transfer> $q$, so the physical absorption process has $p_X=P+q$, $Q^2=-q^2>0$, and $P\cdot q>0$. There is a sign inconsistency in the original PDF: its delta function has $P-p_X-q$, while its positive <Bjorken x> definition and target formulas use incoming $q$. With that literal delta function and $x>0$, the final-state invariant would be
$$
p_X^2=(P-q)^2=M^2-Q^2-2P\cdot q=M^2-Q^2(1+1/x),
$$
which becomes negative in the stated high-energy limit and cannot represent a physical hadronic final state. We therefore use $\delta^4(P+q-p_X)$, preserving the printed $x=Q^2/(2P\cdot q)$ and structure-function conventions. Equivalently, keeping an outgoing momentum variable requires reversing the signs in those definitions as well.
Assume an unpolarized target, incoherent scattering from effectively free spin-one-half <partons>, and the <massless collinear parton approximation>. In the <Bjorken scaling> limit, take $Q^2,P\cdot q\to\infty$ at fixed $0<x<1$, neglecting target and quark masses and intrinsic transverse momentum relative to $Q$. Write the incoming struck-quark momentum as $k=\xi P$ in the leading lightlike-reference approximation. Let $f_f(\xi)$ and $\bar f_f(\xi)$ be quark and antiquark <parton distribution functions>, counting all colours. For the naive scaling result also neglect radiative scaling violations; in full QCD these distributions depend logarithmically on the hard scale.
The outgoing struck quark has momentum $k'=k+q$. Its on-shell condition gives
$$
(k+q)^2=2\xi P\cdot q-Q^2=0,\qquad\boxed{\xi=x.}
$$
Average its initial spin and sum its final spin. The <spin-averaged electromagnetic parton tensor> has numerator
$$
\begin{aligned}
S^{\mu\nu}(k,q)&=\frac12\operatorname{Tr}[\not k\gamma^\mu(\not k+\not q)\gamma^\nu]\\
&=2[k^\mu(k+q)^\nu+k^\nu(k+q)^\mu-g^{\mu\nu}k\cdot q].
\end{aligned}
$$
The supplied three-gamma identity gives this directly: contract its middle index with $(k+q)_\lambda$ and trace with $\not k$; the $\gamma_5$ trace vanishes. Thus no antisymmetric term survives the unpolarized spin average.
Define transverse tensors
$$
T^{\mu\nu}=-g^{\mu\nu}+\frac{q^\mu q^\nu}{q^2},\qquad \widetilde P^\mu=P^\mu-\frac{P\cdot q}{q^2}q^\mu,\qquad \widetilde k^\mu=k^\mu-\frac{k\cdot q}{q^2}q^\mu.
$$
On the cut, $2k\cdot q=Q^2$, so $\widetilde k=k+q/2=\xi\widetilde P$. Expanding the two transverse terms gives
$$
\boxed{S^{\mu\nu}=Q^2T^{\mu\nu}+4\xi^2\widetilde P^\mu\widetilde P^\nu.}
$$
This also explicitly verifies electromagnetic <current conservation>, $q_\mu S^{\mu\nu}=q_\nu S^{\mu\nu}=0$.
The one-particle final-state <Lorentz-invariant phase-space measure> obeys
$$
\int\frac{d^3k'}{(2\pi)^3\,2k'^0}(2\pi)^4\delta^4(k+q-k')=2\pi\delta_+((k+q)^2).
$$
Including the current's charge factor and the given $1/(4\pi)$ normalization, a single quark therefore contributes
$$
w_f^{\mu\nu}(k,q)=\frac{Q_f^2}{2}\delta_+(2k\cdot q-Q^2)S^{\mu\nu}(k,q).
$$
An antiquark has the same unpolarized tensor and the same squared charge $Q_f^2$.
The <parton convolution normalization for a hadronic tensor> is
$$
W^{\mu\nu}(P,q)\simeq\sum_f\int_0^1\frac{d\xi}{\xi}[f_f(\xi)+\bar f_f(\xi)]w_f^{\mu\nu}(\xi P,q).
$$
The $1/\xi$ is important: the distributions count parton numbers, whereas covariantly normalized one-particle responses carry the state normalization $2k^0$. The incoherent sum of normalized responses is $W/(2P^0)=\sum_f\int d\xi(f_f+\bar f_f)w_f/(2k^0)$; in the collinear frame $k^0=\xi P^0$, giving the factor above. There is no extra colour multiplicity because it is already included in the distributions.
On the physical cut,
$$
\delta(2\xi P\cdot q-Q^2)=\frac1{2P\cdot q}\delta(\xi-x).
$$
Comparing the two transverse coefficients in the <hadronic tensor> gives, with $n_f=f_f+\bar f_f$,
$$
W_1\simeq\sum_f\frac{Q_f^2Q^2}{4P\cdot q}\frac{n_f(x)}x=\frac12\sum_fQ_f^2n_f(x),\qquad W_2\simeq\sum_f\frac{Q_f^2}{P\cdot q}\,x n_f(x).
$$
Therefore the requested <deep-inelastic structure functions> are
$$
\boxed{F_1(x,Q^2)\simeq\frac12\sum_fQ_f^2[f_f(x)+\bar f_f(x)],\qquad F_2(x,Q^2)\simeq x\sum_fQ_f^2[f_f(x)+\bar f_f(x)].}
$$
The notation $f_f$ here is the question's $f$ with its flavour index made explicit. The calculation proves the <Callan-Gross relation>, $F_2=2xF_1$, rather than merely citing it. Finite masses and radiative corrections modify this leading approximation; at leading QCD order the same formulas use scale-dependent <parton distribution functions>.
The <quark number sum rule> follows because a distribution is a number density. Integrating $f_f$ counts quarks and integrating $\bar f_f$ counts antiquarks; their difference is the net conserved flavour charge:
$$
\boxed{\int_0^1[f_f(x)-\bar f_f(x)]dx=N_f.}
$$
Every sea quark-antiquark pair contributes zero to this difference. More formally the flavour-vector charge counts quarks minus antiquarks and has a fixed eigenvalue on the hadron state. Thus $N_f$ is the valence or net quark number, not the total number including the sea.
Use both final assumptions printed in the question: only up/down flavours and no antiquarks. The second does not follow just from restricting the flavour list. Then $\int f_u=N_u$, $\int f_d=N_d$, with squared charges $Q_u^2=4/9$ and $Q_d^2=1/9$. The <proton> has $(N_u,N_d)=(2,1)$ and the <neutron> $(1,2)$. Hence the <valence-only proton and neutron structure-function moments> are
$$
\boxed{\int_0^1\frac{dx}{x}F_2^{\rm proton}(x,Q^2)\longrightarrow\frac49\cdot2+\frac19\cdot1=1,\qquad \int_0^1\frac{dx}{x}F_2^{\rm neutron}(x,Q^2)\longrightarrow\frac49\cdot1+\frac19\cdot2=\frac23.}
$$
The integral limits assume the scaling approximation is valid for the indicated weighted moments, for example by <dominated convergence> of the weighted integrands. The same net-quark sum rule alone would not fix these individual moments in the presence of antiquarks, since $F_2$ involves the sum of quark and antiquark distributions.
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