From the $+i\eta$ in the denominator, I'm assuming you're using the $(1,-1)$ signature but then I'm not sure why you wrote $ik\cdot x=-ik^0x^0+ik^1x^1$. More importantly, your second mistake is in the claim of a pole at $k_1-i\eta$. The poles for the $k^0$ integration are at $\pm(|k^1|-i\eta)$, i.e. they involve the magnitude $|k^1|$, not the variable $k^1$ directly.
Besides, it is rather messy to work with several gamma matrices in the integral. A cleaner approach is to push them to the side by writing:
$$
i\int \frac{d^2k}{(2\pi)^2}\frac{\gamma^\mu k_\mu}{k^2+i\eta}e^{ikx}=-i\gamma^\mu\partial_\mu \int \frac{d^2k}{(2\pi)^2}\frac{i}{k^2+i\eta}e^{ikx}
$$
Now you can leave the $-i\gamma^\mu\partial_\mu$ factor alone and work with the integral:
$$
G(x)= \int \frac{d^2k}{(2\pi)^2}\frac{i}{k^2+i\eta}e^{ikx}
$$
You may recognize this as the propagator of a massless scalar. There are already several answers on this site concerning the evaluation of this integral. See, for example, here.
The idea is still the same, depending on the sign of $x^0$ we can choose to close the contour either in the upper or lower plane. But it may be easier to do a Wick rotation and do the integration in Euclidean space. The result is, with the $i\eta$ prescription for $\ln(r)$ omitted :
$$
G(r)=\frac{1}{2\pi}\ln(r)
$$
Here $r=\sqrt{x^2}=\sqrt{x_0^2-x_1^2}$. The result for your original integral is:
\begin{align}
-i\gamma^\mu\partial_\mu \frac{1}{2\pi}\ln(r)&=-\frac{i}{2\pi} \gamma^\mu \frac{1}{r}\partial_\mu \sqrt{x_\mu x^\mu} \\
&= -\frac{i}{2\pi} \frac{\gamma^\mu x_\mu}{x^2}
\end{align}
