If I understand your notation correctly, then your question is a bit confused, because $g^N$ has to be a symmetric matrix, so that "$***$" = "$*$". The condition that $g^N$ is block diagonal does not have to hold; it says that the tangent vector of the last coordinate, $\partial/\partial u^{m+1}$, is perpendicular to the surface $N$. On the other hand, there always exist local coordinates with this property. If you take any local coordinates for $N$, you can evolve them for a short time with the normal surface flow. Indeed you can even get the condition $B = 1$ in a local chart.

Also, there certainly is another way to get the covariant derivative on $M$ and its Christoffel symbol. Namely, if you apply the covariant derivative $\nabla^N$ to a tangent vector field $v$ on $M$ in some tangent direction $w$, you get a vector field $\nabla^N_w(v)$ on $M$ that does not have to be tangent. You should then just project this derivative $\nabla^N_w(v)$ orthogonally onto the tangent bundle $TM$. The orthogonal projection is a useful tensor field $P$ defined on the tangent bundle $TN$ restricted to $M$, and you can write an explicit expression for the covariant derivative $\nabla^M$, or the Christoffel symbol or even the curvature tensor, in terms of $\nabla^N$ and this tensor field $P$. Actually, I am not entirely sure that this method is algebraically all that different, but it is at least conceptually different.

If I understand your notation correctly, then your question is a bit confused, because $g^N$ has to be a symmetric matrix, so that "$***$" = "$*$". The condition that $g^N$ is block diagonal does not have to hold; it says that the tangent vector of the last coordinate, $\partial/\partial u^{m+1}$, is perpendicular to the surface $M$. On the other hand, there always exist local coordinates with this property. If you take any local coordinates for $M$, you can evolve them for a short time with the normal surface flow. You can even get the condition $B = 1$ in a local chart.

Also, there certainly is another way to get the covariant derivative on $M$ and its Christoffel symbol. Namely, if you apply the covariant derivative $\nabla^N$ to a tangent vector field $v$ on $M$ in some tangent direction $w$, you get a vector field $\nabla^N_w(v)$ on $M$ that does not have to be tangent. You should then just project this derivative $\nabla^N_w(v)$ orthogonally onto the tangent bundle $TM$. The orthogonal projection is a useful tensor field $P$ defined on the tangent bundle $TN$ restricted to $M$, and you can write an explicit expression for the covariant derivative $\nabla^M$, or the Christoffel symbol or even the curvature tensor, in terms of $\nabla^N$ and this tensor field $P$. Actually, I am not entirely sure that this method is algebraically all that different, but it is at least conceptually different.