This is about very thin metal layers, much thinner than the skin depth, so it should indeed be obvious that an incoming wave will see as impedance the metal sheet resistance parallel to the wave impedance of the medium behind it. And this parallel impedance is what Eq. 10 is showing. But the authors derive it by beginning with a more general analysis.
As for the question about \$k_2\$, certainly it is a complex number. Since \$k_2\$ is the propagation number in region 2, which is metal, Eq (6) applies, so it is a real number times \$(1+j)\$. Something similar also applies to \$\eta_\text{met}\$, the wave impedance in the metal. More importantly, this \$\eta_\text{met}\$ is very low, because the metal as a conductor has a very low (and lossy) wave impedance. We further need the fact that the sheet resistance is:
$$ R_\text{s}=\frac{1}{\sigma\,d} $$
I don't see in your picture how the authors defined \$\eta_\text{met}\$, but it should be somewhere and it should be similar to Eq. (6) for \$k_2\$. If you divide those two quantities the simple result is:
$$
\frac{\eta_\text{met}}{k_2} =\frac{-j}{\sigma} \tag{1}
$$
Now we are ready to tackle Eq. 8:
$$\begin{align}
Z_{1-2}(d) = \eta_\text{met}\ \frac{\eta_0-j\,\eta_\text{met}\tan(k_2d)}
{\eta_\text{met}-j\,\eta_0\tan(k_2d) }
\approx \eta_\text{met}\ \frac{\eta_0-j\,\eta_\text{met}\,k_2d}
{\eta_\text{met}-j\,\eta_0\ k_2d}
\\[10pt]
= \ \frac{\eta_0\,\eta_\text{met}/(k_2d)-j\,\eta_\text{met}^2}
{\eta_\text{met}/(k_2d)-j\,\eta_0 }
= \ \frac{-\eta_0\,j/(\sigma d)-j\,\eta_\text{met}^2}
{-j/(\sigma d)-j\,\eta_0 }
\end{align}$$
where the second line used result \$(1)\$ from above and the first line uses the approximation of Eq. (7) in the paper. Now neglect the term with \$\eta_\text{met}^2\$ because it is small, and use the definition of \$R_s\$:
$$\begin{align}
Z_{1-2}(d) &= \ \frac{-\eta_0\,j/(\sigma d)-j\,\eta_\text{met}^2}
{-j/(\sigma d)-j\,\eta_0 }
\approx \ \frac{-\eta_0\,j/(\sigma d)}
{-j/(\sigma d)-j\,\eta_0 }
= \ \frac{\eta_0\, R_\text{s}}
{\eta_0+ R_\text{s} }
\end{align}$$