Consider a manifold $M^2\subset\mathbb{R}^4$ locally parametrized by $Y(\xi^1,\xi^2)$ and let $X$ be a Fermi coordinates parametrization of a neighbourhood of $M$ (which we call $\bar{M}$), that is $$ X(\xi^1,\xi^2,z^1,z^2)=Y(\xi^1,\xi^2)+z^1\nu_1(\xi^1,\xi^2)+z^2\nu_2(\xi^1,\xi^2) $$ where $\nu_1,\nu_2$ are two orthonormal fields. In the codimension 1 case it's fairly easy to see that the metric tensor for $\bar{M}$ is given by $$ g_{\bar{M}}=\begin{bmatrix}g_{M_{z}} & 0\\ 0 & 1 \end{bmatrix} $$ where $M_z=M+z_1\nu_1+z_2\nu_2$. In codimension 2 I expect the same block decomposition for the metric tensor $$ g_{\bar{M}}=\begin{bmatrix}g_{M_{z}} & 0\\ 0 & I_2 \end{bmatrix} $$ but I can't prove it through rough calculations. In short, the problem is given by the fact that, in general $$ \partial_j\nu_1\cdot\nu_2=-\partial_j\nu_2\cdot\nu_1\ne0 $$ and this nonzero term gives a nonzero element on the anti-diagonal blocks. Hence, this block decomposition actually depends on the choice of $\nu_1,\nu_2$ somehow.
Here Existence of two normal vector fields with derivatives lying in $TM$ is described how I tried to "rotate" the basis for $TM^\perp$ to obtain another one with $ \partial_j\nu_1\cdot\nu_2=-\partial_j\nu_2\cdot\nu_1=0 $ but this didn't yield to any conclusion. Is there any quicker way to ensure the existence of the block decomposition?
Any help is very appreciated.