I think the BC you are looking for this boundary is the following: $$ \vec{n}^T\sqrt{g^{\alpha\beta}}\left[\vec{grad}(\phi)\right]=0$$ Where $g^{\alpha\beta}$ is the contravariant metric tensor, $\vec{n}$ is the exterior normal vector to upper and lower boundaries and the gradient operator is expressed in the new coordinates.

EDIT Since in a symmetry plane no flow crosses this boundary we have that the BC is:

$$\vec{n}^T\vec{u}=0$$ Therefore, since $\vec{u}=\vec{grad}\,\phi$ you have the required zero normal gradient BC $$\vec{n}^T\vec{grad}\,\phi=0$$

IMPORTANT

To keep things clear... the Laplacian operator $\triangle$ must come with ONLY one BC in each point of the boundary, this is, you can not specify 2 BC as you propose on the boundary. To se this.. I ask you sth. How many BC you need in the boundary to solve this unidimensional BVP? $$\frac{d^2 u}{d x^2}=0,\quad x\in[x_1,x_2]$$

I think the BC you are looking for this boundary is the following: $$ \vec{n}^T\sqrt{g^{\alpha\beta}}\left[\vec{grad}(\phi)\right]=0$$ Where $g^{\alpha\beta}$ is the contravariant metric tensor, $\vec{n}$ is the exterior normal vector to upper and lower boundaries and the gradient operator is expressed in the new coordinates.

EDIT Since in a symmetry plane no flow crosses this boundary we have that the BC is:

$$\vec{n}^T\vec{u}=0$$ Therefore, since $\vec{u}=\vec{grad}\,\phi$ you have the required zero normal gradient BC $$\vec{n}^T\vec{grad}\,\phi=0$$

I think the BC you are looking for this boundary is the following: $$ \vec{n}^T\sqrt{g^{\alpha\beta}}\left[\vec{grad}(\phi)\right]=0$$ Where $g^{\alpha\beta}$ is the contravariant metric tensor, $\vec{n}$ is the exterior normal vector to upper and lower boundaries and the gradient operator is expressed in the new coordinates.

I think the BC you are looking for this boundary is the following: $$ \vec{n}^T\sqrt{g^{\alpha\beta}}\left[\vec{grad}(\phi)\right]=0$$ Where $g^{\alpha\beta}$ is the contravariant metric tensor, $\vec{n}$ is the exterior normal vector to upper and lower boundaries and the gradient operator is expressed in the new coordinates.

EDIT Since in a symmetry plane no flow crosses this boundary we have that the BC is:

$$\vec{n}^T\vec{u}=0$$ Therefore, since $\vec{u}=\vec{grad}\,\phi$ you have the required zero normal gradient BC $$\vec{n}^T\vec{grad}\,\phi=0$$