Here is my attempt. But I'm not sure if it is right.
Based on the result, I could guess what's might be the right way to get the answer. For simplicity, we only consider the $C_1$ integral. We first replace the slowly varying term by their value on the real axis. The term colored red is fast varying(and divergent when replacing $q$ by $2k_F + i \eta$)
$$
\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} \int_{C_1} q \mathrm{d}q e^{iqx}\left(q^2\left\{q^2 + q_{TF}^2\left[\frac{1}{2} - \frac{k_F}{2q}(1-\frac{q^2}{4k_F^2})\frac{1}{2}\ln \frac{(q-2k_F)^2 + \eta^2}{(q+2k_F)^2 + \eta^2}\right]\right\}^{-1} - 1\right)\\
=\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} \int_{\eta}^{\infty} 2k_F i\mathrm{d}v e^{2ik_F x} e^{-v x} \Delta\left((2k_F)^2 \left\{(2k_F)^2 + q_{TF}^2 \left[\frac{1}{2} - \frac{1}{4}\color{red}{(1-\frac{q^{2}}{4k_F^2})\frac{1}{2}\ln \frac{(q-2k_F)^2 + \eta^2}{(q + 2k_F)^2 + \eta^{2}}}\right]\right\}^{-1} \right)\\
=\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} e^{2ik_F x}\int_{\eta}^{\infty} i \mathrm{d}v e^{-vx}(2k_F)^3 \Delta\left\{\left[(2k_F)^2 + \frac{1}{2}q_{TF}^2 \right]- \frac{1}{4}q_{TF}^2(1-\frac{q^{2}}{4k_F^2})\frac{1}{2}\ln \frac{(q-2k_F)^2 + \eta^2}{(q + 2k_F)^2 + \eta^{2}}\right\}^{-1}
$$
The $\Delta$ operator represents the difference of two sides of the branch cut.
Then we treat the term $[\ldots]$ as the dominant term and Taylor expand the denominator(See the justification below):
$$
=\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} e^{2ik_F x}\int_{\eta}^{\infty} i \mathrm{d}v e^{-vx}(2k_F)^3 \Delta\left\{\frac{1}{(4k_F^2 + \frac{1}{2}q_{TF}^2)^2}\frac{1}{4}q_{TF}^2(1-\frac{q^{2}}{4k_F^2})\frac{1}{2}\ln \frac{(q-2k_F)^2 + \eta^2}{(q + 2k_F)^2 + \eta^{2}}\right\}\\
=\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} e^{2ik_F x}\int_{\eta}^{\infty} i \mathrm{d}v e^{-vx}(2k_F)^3 \frac{1}{(4k_F^2 + \frac{1}{2}q_{TF}^2)^2}\frac{1}{4}q_{TF}^2(1-\frac{q^{2}}{4k_F^2}) \Delta\left[\frac{1}{2}\ln \frac{(q-2k_F)^2 + \eta^2}{(q + 2k_F)^2 + \eta^{2}}\right]
$$
The zeroth order term in the expansion vanish due to the $\Delta$ operator.
Now, since $\Delta\left[\frac{1}{2}\ln \frac{(q-2k_F)^2 + \eta^2}{(q + 2k_F)^2 + \eta^{2}}\right] = i\pi$ on contour $C_1$, we have:
$$
=\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} e^{2ik_F x}\int_{\eta}^{\infty} i \mathrm{d}v e^{-vx}\frac{(2k_F)^3}{(4k_F^2 + \frac{1}{2}q_{TF}^2)^2}\frac{1}{4}q_{TF}^2(1-\frac{q^{2}}{4k_F^2}) i\pi\\
=\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} e^{2ik_F x}\int_{\eta}^{\infty} i \mathrm{d}v e^{-vx} \frac{(2k_F)^3}{(4k_F^2 + \frac{1}{2}q_{TF}^2)^2}\frac{1}{4}q_{TF}^2(1-\frac{(2k_F+iv)^{2}}{4k_F^2}) i\pi\\
=\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} e^{2ik_F x}\int_{\eta}^{\infty} i \mathrm{d}v e^{-vx} \frac{(2k_F)^3}{(4k_F^2 + \frac{1}{2}q_{TF}^2)^2}\frac{1}{4}q_{TF}^2\frac{-iv}{k_F^2} i\pi\\
=\frac{Ze}{4\pi^2 i x} \lim_{\eta \to 0} e^{2ik_F x}\frac{(2k_F)^3}{(4k_F^2 + \frac{1}{2}q_{TF}^2)^2}\frac{\pi q_{TF}^2}{4k_F^2} \int_{\eta}^{\infty} i v e^{-vx} \mathrm{d}v
$$
where in the third line we ignore the $v^2$ term due to the decaying term $e^{-vx}$.
For the justification of Taylor expansion. Maybe one possible explanation is that, we can break the integration into two part:$\int_{2\eta}^{\infty} + \int_{\eta}^{2\eta}$.
For the second part:
$$
\int_{\eta}^{2\eta} i \mathrm{d}v e^{-vx}(2k_F)^3 \Delta\left\{\left[(2k_F)^2 + \frac{1}{2}q_{TF}^2 \right]- \frac{1}{4}q_{TF}^2(1-\frac{q^{2}}{4k_F^2})\frac{1}{2}\ln \frac{(q-2k_F)^2 + \eta^2}{(q + 2k_F)^2 + \eta^{2}}\right\}^{-1}
$$
Notice that $\ln[(q-2k_F)^2+\eta^2]$ diverge when $v \to \eta^+$, but the divergence is at the denominator, thus the integrand is bounded in this region, thus this part of the integral is at most $\mathcal{O}(\eta)$. In the final limiting process, this extra contribution can be ignored.
For the first part, again, since $x \to \infty$, we only need to care about the integrand at small $v$, but for $v = 2\eta$, $(1-\frac{q^{2}}{4k_F^2})\ln[(q-2k_F)^2+\eta^2] = -2i\eta\ln[3\eta^2]/k_F + \mathcal{O}(\eta^2)$. This term is indeed small for a small $\eta$. Thus the Taylor expansion is justified.
The final result is not changed, we just replace $\int_{\eta}^{\infty} i v e^{-vx} \mathrm{d}v$ to $\int_{2\eta}^{\infty} i v e^{-vx} \mathrm{d}v$ but since we take $\eta \to 0$, the result is the same.
Maybe my attempt seems 'ugly'. Is it right? Are there some more elegant approach?
