I think you have not intepreted the question correctly (or at least you have left too many words out that I cannot tell whether you have interpreted it correctly).
In particular, I do not see that you have actually used the definition of a smooth manifold, in particular you have not made use of the key fact about smooth manifolds, namely smoothness of overlap maps. You write about two coordinate systems $x$ and $y$, and you mention the overlap $x^{-1} y$, but you never mention or make use of the key fact that the overlap maps $x^{-1} y$ and $y^{-1} x$ are smooth.
Let me separate the definition of a Riemannian metric on $M$ into two parts:
- The object given;
- The property that object must satisfy.
The object given is:
The correspondence which associates to each point $p \in M$ an inner product $\langle,\rangle_p$ on the tangent space $T_p M$".
Up to here, coordinate systems are not involved, but now they come into the property that the correspondence must satisfy:
If $x:U\subset\mathbb R^n\rightarrow M$ is a coordinate system around $p$, with $x(x_1,...,x_n)=q\in x(U)$, then $g_{i,j}(x_1,...,x_n):=<dx_q(e_i),dx_q(e_j)>_q$ is a smooth function on $U$.
The issue is whether this property depends on the coordinate system.
What one must show is if $y:V\rightarrow M$ is another coordinate system around $p$, and if we let $h_{i,j}(y_1,...,y_n) := \langle dy_q(e_i),dy_q(e_j) \rangle_q$, then on the overlap $U \cap V$, to say that each of the functions $g_{i,j} \mid x(U \cap V)$ is smooth is equivalent to saying that each of the functions $h_{i,j} \mid y(U \cap V)$ is smooth.
The way you prove this equivalence is to write out how to relate the maps $g_{i,j} \mid x(U \cap V)$ and $h_{i,j} \mid y(U \cap V)$ (for each $i,j$) are related, expressing this relation as a equation obtained from the chain rule, involving the Jacobian matrices for the overlap maps $x^{-1} y$ and $y^{-1} x$.
Smoothness is a local property, so consider an open neighborhood $U\subseteq M$ of $p \in M$, and extend $X$ and $Y$ to vector fields $\widetilde{X}$ and $\widetilde{Y}$ on $U$. So the function $U \cap N \ni q \mapsto \langle X_q,Y_q\rangle_N$ is the restriction of the smooth map $U \ni q \mapsto \langle \widetilde{X}_q,\widetilde{Y}_q\rangle_M$, which by assumption is smooth. Restrictions of smooth maps are smooth, and so you are done.
If you're a more coordinate-oriented person, you can take coordinates $(x^j, y^\mu)$ adapted to $N$, i.e., such that $N$ is described by $y^\mu = 0$, so that the tangent spaces to $N$ are spanned by the first coordinate fields $\partial_j$. Then the metric matrix has the block form $$\begin{pmatrix} (g_{jk}) & (g_{j\lambda}) \\ (g_{\lambda k}) & (g_{\mu\nu})\end{pmatrix}$$where all the entries are smooth. In particular the first block $(g_{jk})$ is smooth, so you're done.
Best Answer
No, this does not work, because this definition is not independent of the parametrization chosen. If you pick two different parametrizations $x:U\to M$ and $y:V\to M$ at the same point $q\in M$, they will usually define different metrics on the tangent space at $q$ (essentially because $dx_q^{-1}\circ dy_q$ will usually not preserve the dot product, since it could be any invertible linear map).
To get a well-defined metric on all of $M$ instead of just on one coordinate chart, you have to combine these "dot products" with a partition of unity. Namely, take an open cover of $M$ by coordinate charts and use this "dot product metric" on each one, and then add them all up to get a metric on all of $M$ using a partition of unity subordinate to the cover. This of course will be highly noncanonical since the resulting metric on $M$ you get depends on the choice of coordinate charts and partition of unity.