Let $$B=\{(x,y,z) \in R^3 : x^2+y^2=4z\}$$
$$L=\{(x,y,z) \in R^3 : x=2, y=0\}$$
Let $P$ be a fragment of $B$ of points such that distance from them to $L$ is less than 1. $P$ is oriented as follows: in any point, normal vector giving positive side has positive $z$ component.
Calculate flux over vector field $F(x,y,z)=(\frac{-y}{z},\frac{x}{z}, \frac{y}{x-z})$ through $P$ from negative to positive side.
So I calculated that $P=\{(x,y,z) \in R^3 : x^2+y^2=4z, (x-2)^2+y^2<1\}$ and that normal vector giving positive side points inwards. I considered parametrization $f(u,v)=(u,v,\frac{u^2+v^2}{4})$ from $\operatorname{Ball} K=\{(u-2)^2+v^2<1\}$ to $P$.
So:
$$\operatorname{Flux} = \int_P\frac{-y}{z}dy \wedge dz -\frac{x}{z} dx \wedge dz + \frac{y}{x-z} dx \wedge dy=(\color{red}{\pm1})\int_K \frac{-v}{\frac{u^2+v^2}{4}}(\frac{-u}{2})-\frac{u}{v}(\frac{v}{2})+\frac{v}{u-\frac{u^2+v^2}{4}}\qquad (1)$$
and I am not sure about the sign after equality (in red). I think it should be negative because normal point inwards, can you explain it and check if everything before is correct?
My previous answer was too involved and came to the wrong conclusion for some reason.
The answer is you need the plus sign.
The surface $B$ is the boundary of an elliptic paraboloid. I gather that $z\ge 0\,.$ The inward pointing unit normal vector at $B$ is $$\tag{1} \mathbf{n}=\frac{1}{\sqrt{x^2+y^2+4}}\begin{pmatrix}-x\\-y\\2\end{pmatrix}\,. $$ Following Ted Shifrin's comment: The normal vector is $$\tag{2} \frac{\partial f}{\partial u}\times \frac{\partial f}{\partial v}=\begin{pmatrix}1\\0\\\frac{u}{2}\end{pmatrix}\times \begin{pmatrix}0\\1\\\frac{v}{2}\end{pmatrix}=\begin{pmatrix}-\frac{u}{2}\\-\frac{v}{2}\\\color{red}{+}1\end{pmatrix} $$ when surface parametrized by $f(u,v)=(u,v,(u^2+v^2)/4)^\top\,.$ The normal vector (2) has positive $z$-component and points inward like $\mathbf{n}$ does.
The flux of $F$ through $P$ in the direction of those normal vectors is $$ \int_PF\cdot \mathbf{n}\,dS= \int_K F\cdot\Big\{\frac{\partial f}{\partial u}\times \frac{\partial f}{\partial v}\Big\}\,du\,dv\,.\tag{3} $$ (The reason why the cross product does not need to be normalized is discussed in this answer.)
To bring (3) into a notation using differential forms we write $f=(x,y,z)^\top\,.$ Then the components of $\{\frac{\partial f}{\partial u}\times \frac{\partial f}{\partial v}\}\,du\,dv $ are \begin{align} \Big\{\frac{\partial y}{\partial u}\frac{\partial z}{\partial v}-\frac{\partial z}{\partial u}\frac{\partial y}{\partial v}\Big\}\,du\,dv &=dy\wedge dz\,,\\ \Big\{\frac{\partial z}{\partial u}\frac{\partial x}{\partial v}-\frac{\partial x}{\partial u}\frac{\partial z}{\partial v}\Big\}\,du\,dv &=dz\wedge dx\,,\\ \Big\{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\Big\}\,du\,dv &=dx\wedge dy\,. \end{align} Therefore with the desired inward orientation of $\mathbf{n}$ the flux is $$ \int_PF\cdot \mathbf{n}\,dS=\int_PF_x\,dy\wedge dz+F_y\,dz\wedge dx+F_z\,dx\wedge dy\,. $$ In your case this is calculated by the RHS of (3) and using (2) as $$ \int_K\frac{-v}{\frac{u^2+v^2}{4}}(\frac{-u}{2})-\frac{u}{\color{red}{\frac{u^2+v^2}{4}}}(\frac{v}{2})+\frac{v}{u-\frac{u^2+v^2}{4}}\,du\,dv $$ where I fixed an error of yours.