1. This is a surface current, so it would be represented by [itex]\vec{K}[/itex] not [itex]\vec{J}[/itex].
2. How did [itex]\int\vec{J} \cdot d\vec{s}[/itex] turn into [itex]\int\vec{J} \cdot d\vec{l}[/itex]. [itex]d\vec{l}[/itex] is a line element around the circuit, while [itex]d\vec{s}[/itex] is an area element of the enclosed surface. You can't turn one into the other.
3. It helps to define a coordinate system, this way you can loose [itex]d\vec{l}[/itex] in favor of [itex]\hat{x}dx[/itex], [itex]\hat{y}dy[/itex], or [itex]\hat{z}dz[/itex], and the same for [itex]d/vec{s}[/itex]. This is will avoid confusion when trying to determine [itex]\vec{B}\cdot d\vec{l}[/itex]. [itex]d/vec{l}[/itex] is pointing in the opposite direction along [itex]C_2[/itex], not that
4. This is the most important part. The way you drew the loop, [itex]\vec{J}\cdot d/vec{s}=0[/itex]. What that tells you is that there is no field parallel to the current.
5. The way you should have drawn the loop is so that the loop is perpendicular to the current. Then [itex]\vec{J}\cdot d\vec{s}=Kda[/itex] and [itex]\int\vec{J}\cdot d\vec{s}=Kl[/itex]. This tells you that the field is perpendicular to the surface current.
6. You can't assume that the field is parallel to the plane, so the C2 and C3 integrals aren't necessarily 0. What happens is that the fields are the same along both lines, but because they are in opposite directions, they cancel.
I have a few more things to say, but my laptop is running out of power, I'll continue in the morning.