@Tamir: what book are you learning from? It seems that you are missing/confounding some elementary concepts.

This really belongs as a clarification of my earlier comment, but it doesn't look like it will fit in the comment box. Hence now it lives as an answer.

To start with we set some notations.

$(N,g^N)$ is a Riemannian manifold. This means that $N$ is a smooth manifold (locally diffeomorphic to domains in $\mathbb{R}^{m+1}$) and $g^N|_p$ at some point $p\in N$ is a positive definite (and hence non-degenerate) symmetric bilinear form on $T_pN$. So given an arbitrary vector $v\in T_pN$, $g^N(v)$ is a co-vector, or an element of $T^*_pN$. $T_pN$ and $T_p^*N$ are *not* the same space; the metric however induces a canonical isomorphism between the two (since the metric can be understood as a non-degenerate map between the two vector spaces).

Now given $M$ a $m$-dimensional smooth submanifold of $N$. The identity map $\iota:M\to N$ is an embedding. The tangent space $T_pM$ for $p\in M$ naturally embeds into $T_{p}N$ by the tangent bundle map $d\iota: TM\to TN$. (Note that, however, without the Riemannian structure there's no natural embedding of the dual space $T^*_pM$ into $T_p^*N$; but don't worry about that now, since we won't need it.)

Now I give an explicit construction of a coordinate system in a neighborhood $U$ of $p$ such that the metric $g^N$ is block diagonal along points $q \in U\cap M$.

First fix a neighborhood $V\subset M$ of $p$, and an arbitrary coordinate system $u^1,\ldots,u^m$ on $V$, with $p$ at the origin. At every point $q\in V$ the vectors $\partial_i = \partial/\partial u^i$ span the tangent space $T_qM$. Now consider the image of $\partial_i$ under the map $d\iota$, call them $e_i = d\iota \partial_i$. By elementary linear algebra since $T_qM\subset T_qN$ has co-dimension 1, there exists a unique vector $n_q$, up to scaling, such that $g(n_q,e_i) = 0$ for all $i$. (The reason I used $e_i$ instead of $\partial_i$ on $T_qN$ is because without a full set of coordinates [we're still missing one] it doesn't make sense to speak of the "coordinate derivative", which is obtained by varying one coordinate value while holding the remainder fixed.)

In any case, so in $T_qN$ for any point $q\in M$, we now have a set of basis vectors $e_1,\ldots,e_m,n_q$. Normalize $n_q$ so that $g(n_q,n_q) = 1$ and $\eta(e_1,e_2,\ldots,e_m,n_q) > 0$ where $\eta$ is the volume form on $N$. (So now we fix the size and orientation of the field $n_q$.) Now we extend $n_q$ somehow: pick your favourite way. One possible way is to extend $n_q$ by the geodesic flow for some short period of time. Let's call $\gamma(t,q)$ the geodesic starting from the point $q$ with initial speed $n_q$ at the time $t$. By possibly shrinking the neighborhood $V$, there exists some $\epsilon > 0$ such that $\gamma: (-\epsilon,\epsilon)\times V \to N$ is injective.

Now let the neighborhood $U = \gamma( (-\epsilon,\epsilon)\times V )$. Define the coordinate on $U$ thus: for any point $x\in U$, let $\tilde{u}^i(x) = u^i\circ \pi_2\circ\gamma^{-1}(x)$, where $\pi_2: (-\epsilon,\epsilon)\times V \to V$ is projection onto the second component. In other words, at a point $x$, find the unique geodesic $\gamma(t,q)$ that passes through $x$, and set the value $\tilde{u}^i(x) = u^i(q)$. To complete the coordinate system you let $\tilde{u}^{m+1}(x) = \pi_1\circ \gamma^{-1}(x)$, where $\pi_1$ is projection onto the first component, or that if $\gamma(t,q)$ is the geodesic, set $\tilde{u}^{m+1}(x) = t$.

Now you simply check that by definition, the surface $\{t = 0\} \cap U = V$. And that along this surface $\partial/\partial \tilde{u}^{m+1} = n_q$, and $\partial/\partial \tilde{u}^i = e_i$ for the other $i$'s. So that for any $q\in V$ the metric $g^N$ is block diagonal. Furthermore, observe that
$$ \partial_{m+1} g(\partial_{m+1},\partial_{m+1}) = 2 g(\partial_{m+1},\nabla_{\partial_{m+1}}\partial_{m+1}) = 0$$
by the geodesic equation, you have that $g(\partial_{m+1},\partial_{m+1}) = 1$ on $U$. Also, use that
$$ \partial_{m+1} g(\partial_i,\partial_{m+1}) = g(\nabla_{\partial_{m+1}}\partial_i,\partial_{m+1}) = g(\nabla_{\partial_i}\partial_{m+1},\partial_{m+1}) = \frac12 \partial_i g(\partial_{m+1},\partial_{m+1}) = 0$$
(first equality uses the geodesic equation again, the second one uses that Levi-Civita connection is torsion free, and the Lie bracket of coordinate vector fields vanish). We see that $g(\partial_i,\partial_{m+1}) = 0$ on $U$. So in the whole neighborhood $U$, we have that $g^N$ is block diagonal, with the block $B$ exactly 1.