= Solution
The local matrix $U(x)$ is a <gauge transformation> in the <representation> acting on the matter field. With Hermitian <Lie algebra generators> it can be written locally in spacetime, and pointwise for $SU(3)$, as
$$
U(x)=\exp[i\omega^a(x)T^a],\qquad [T^a,T^b]=if_{abc}T^c.
$$
The real functions $\omega^a$ can instead include a factor of $g$ by redefining the transformation parameters. The <gauge covariant derivative> convention in this paper is $D_\mu=\partial_\mu+igA_\mu$. Inserting the stated transformed $A_\mu$ shows explicitly that the $\partial_\mu U$ term cancels, giving $D_\mu'(U\psi)=UD_\mu\psi$.
In <QCD>, each <quark field> is in the colour <fundamental representation> $\mathbf3$, with $\psi^i\mapsto U^i{}_j\psi^j$. Thus its colour generators are \b[eight $3\times3$ matrices], conventionally $T^a=\lambda^a/2$ for the <Gell-Mann matrices>; their number is not their matrix dimension. The eight <gluons> themselves transform in the <adjoint representation of SU(3)>.
With $\operatorname{tr}(T^aT^b)=\delta^{ab}/2$, the pure <Yang-Mills theory> has Lagrangian $\mathcal L_{\mathrm{YM}}=-\tfrac14F^a_{\mu\nu}F^{a\mu\nu}=-\tfrac12\operatorname{tr}(F_{\mu\nu}F^{\mu\nu})$. Put $G^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu$ and $C^a_{\mu\nu}=f_{abc}A^b_\mu A^c_\nu$. Since $F=G-gC$,
$$
\mathcal L_{\mathrm{YM}}=-\frac14G^a_{\mu\nu}G^{a\mu\nu}+\frac g2G^a_{\mu\nu}C^{a\mu\nu}-\frac{g^2}4C^a_{\mu\nu}C^{a\mu\nu}.
$$
The cross term has two derivative contributions. Interchange $\mu,\nu$ and then $b,c$ in the second; antisymmetry of $f_{abc}$ makes it equal to the first. The <cubic and quartic Yang-Mills self-interactions> are therefore
$$
\boxed{\mathcal L_3=gf_{abc}(\partial_\mu A^a_\nu)A^{b\mu}A^{c\nu},\qquad
\mathcal L_4=-\frac{g^2}4f_{abc}f_{ade}A^b_\mu A^c_\nu A^{d\mu}A^{e\nu}.}
$$
The positive cubic sign follows from the negative $gfAA$ term in the given <Yang-Mills field strength>; changing the covariant-derivative convention consistently changes intermediate signs.
For the left-handed <quark doublet> $Q_L=(u_L,d_L)^T$, the <hypercharge> is $Y=1/6$ and the weak generators are $\sigma^\alpha/2$. Its electroweak term is
$$
\boxed{\mathcal L_{Q_L}=\overline Q_L i\gamma^\mu\left[\partial_\mu+\frac{ig}{2}\sigma^\alpha\mathcal A_\mu^\alpha+\frac{ig'}6 B_\mu\right]Q_L.}
$$
The <Pauli matrices> give diagonal entries $\pm\mathcal A^3$ and off-diagonal entries $\mathcal A^1\mp i\mathcal A^2$. Expanding the term gives
$$
\begin{aligned}
\mathcal L_{Q_L}={}&\bar u_Li\gamma^\mu\partial_\mu u_L+\bar d_Li\gamma^\mu\partial_\mu d_L\\
&-\frac{g'}6B_\mu(\bar u_L\gamma^\mu u_L+\bar d_L\gamma^\mu d_L)
-\frac g2\mathcal A^3_\mu(\bar u_L\gamma^\mu u_L-\bar d_L\gamma^\mu d_L)\\
&-\frac g2\left[(\mathcal A^1_\mu-i\mathcal A^2_\mu)\bar u_L\gamma^\mu d_L
+(\mathcal A^1_\mu+i\mathcal A^2_\mu)\bar d_L\gamma^\mu u_L\right].
\end{aligned}
$$
In particular the off-diagonal <weak charged current> coupling is $-g(W_\mu\bar u_L\gamma^\mu d_L+W_\mu^*\bar d_L\gamma^\mu u_L)/\sqrt2$.
Write $s_W=\sin\theta_W$, $c_W=\cos\theta_W$. Invert the neutral-field rotation: $\mathcal A^3_\mu=c_WZ_\mu+s_WA_\mu$, $B_\mu=c_WA_\mu-s_WZ_\mu$. The original PDF gives $\tan\theta_W=g'/g$, so
$$
\boxed{e=g\sin\theta_W=g'\cos\theta_W.}
$$
The <photon> coefficients inside the derivative are $gs_W/2+g'c_W/6=2e/3$ for $u_L$, and $-gs_W/2+g'c_W/6=-e/3$ for $d_L$. Multiplication by the $i$ in the kinetic Lagrangian gives the requested <electromagnetic coupling of a weak multiplet>:
$$
\boxed{\mathcal L_{\mathrm{em}}=-\frac23e\bar u_LA_\mu\gamma^\mu u_L+\frac13e\bar d_LA_\mu\gamma^\mu d_L.}
$$
Equivalently it is $-eA_\mu\overline Q_L\gamma^\mu(T_3+Y)Q_L$, making the <electric charges> explicit. The remaining neutral terms are $-(g/c_W)Z_\mu\{(1/2-2s_W^2/3)\bar u_L\gamma^\mu u_L+(-1/2+s_W^2/3)\bar d_L\gamma^\mu d_L\}$. The TeX's missing prime in the mixing-angle ratio and truncated first Pauli matrix are conversion errors, not physical assumptions.
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