Context:
In Cartesian coordinates, consider a vector field $\boldsymbol{f}=\boldsymbol{f}(x,y)$ defined over some domain $\Omega$ in the $xy$ plane. Taking a closed contour $\Gamma$ that encloses surface area $A$ at any point inside the same domain ($\Omega$), then we can define the two integrals
$$I_{1}=\int_{\Gamma}\boldsymbol{f}\cdot\boldsymbol{\hat{n}}dl,$$ $$I_{2}=\int_{\Gamma}\boldsymbol{f}\cdot\boldsymbol{dl},$$
where $dl$ is the length element along the contour, $\hat{n}$ is the unit vector normal (perpendicular) to the element $dl$ at each point in the plane, and bold symbols represent vector quantities.
To make $I_{2}=0$, we can choose the contour $\Gamma$ shape to be orthogonal to the field flowlines at each point, whereas to make $I_{1}=0$, we choose them to be parallel at each point. For example, if the field is generated radially from a point source in the plane, we can choose the contour shapes to be concentric circles around the source point, so that these circles will always cross the radial field lines perpendicularly and give $I_{2}=0$. Or if the field is circular flow, we can choose the contour to also be circular, and thus get $I_{1}=0$. These examples should work regardless of how large/small the enclosed area $A$ is taken.
The question:
Is there any nontrivial example of a family of fields and contour shapes that we can choose that will give us $I_{1}=0$ or $I_{2}=0$ at every point (not necessarily concentric) and for any size of enclosed area $A$ (including when $A\rightarrow0 $) in the domain of the problem? Can you given any practical examples of scenarios that could lead to this?
And if this is impossible to satisfy at every point, is there a theorem/proof about this?
It seems like you don't know Green's Theorem (or, more generally, Stokes's Theorem and Gauss's Theorem, also called the Divergence Theorem). Assuming your vector field is continuously differentiable, you certainly can't even accomplish your goal for one point unless you have strong conditions on $\text{div}\, \boldsymbol f$ and $\text{curl}\,\boldsymbol f$. To get $I_1=0$ in a non-instantaneous fashion, you'll need $\text{div}\, \boldsymbol f=0$ on a region containing your point, and, likewise, to get $I_2=0$ you'll need $\text{curl}\,\boldsymbol f = \mathbf 0$ on a region. The latter happens, in particular, for conservative force fields, i.e., when $\boldsymbol f = \nabla\phi$ for some function $\phi$. [Geometrically, you can't get curves as you described unless the field flowlines form closed curves; usually that won't happen. Nor, in general, will the orthogonal trajectories of the field lines form closed curves. The examples you have in your mind are special cases indeed.]
Side remark: You can get vanishing of both $I_1$ and $I_2$ when $\boldsymbol f = \nabla\phi$ and $\phi$ is harmonic on your region in the plane. Then both the curl and divergence will vanish.