The 4 conditions above are what I am being taught, and I am aware of the usual boundary conditions of electric field and potential, but that links to surface charge density, and I was wondering how do I solve it in 2D case, where in this question, they ask for linear charge density, how should I link the boundary conditions equations to get linear charge density?
In the boundary condition $$ \frac{\partial {V_{above}}}{\partial {n}}-\frac{\partial {V_{below}}}{\partial {n}}=-\frac{\sigma}{\epsilon_0} $$ you have $V_{above}=0$ and the normal points in the $y$ direction. $$ \sigma=\epsilon_0\frac{\partial {V}}{\partial {y}} $$
Summary: just replace $\sigma$ with $\lambda$ and everything's the same otherwise.
In 2-D, the integral form of Gauss's Law is still $$ \frac{Q_\text{enc}}{\epsilon_0} = \oint \vec{E} \cdot d\vec{a}. $$ but now the integral is over a closed loop (rather than a surface, with $d\vec{a}$ and $Q_\text{enc}$ refers to the charge enclosed in an area (not a volume).
To figure out how this relates to the discontinuity in $V$ at a boundary, we imagine taking as our integration volume a narrow rectangle, with two longer sides of length $\ell$ running parallel to the the boundary at a distance $\pm \epsilon$ inside & outside, and two short sides connecting the ends of the longer sides. In other words, we take the 2D analog of a "Gaussian pillbox". By the same logic as in the 3D case, we have $$ \frac{\lambda \ell}{\epsilon_0} = \ell(\vec{E}_1 \cdot \vec{n}) - \ell (\vec{E}_2 \cdot {n}) $$ where $\hat{n}$ is a unit normal pointing from region 2 into region 1. The factors of $\ell$ cancel, and since we have $\vec{E}_1 \cdot \hat{n} = - \partial V_1/\partial n$, we conclude that $$ - \frac{\lambda}{\epsilon_0} = \frac{\partial V_1}{\partial n} - \frac{\partial V_2}{\partial n}, $$ just as we would expect from naïvely replacing $\sigma \to \lambda$.
