Suppose $M$ is a manifold of dimension $n$.Let $p\in M$ and let $\psi:U\subset \mathbb R^n\to M$ be a local parametrization on $M$ near $p$.So,$\psi:U\to \psi(U)$ where $\psi(U)$ is open in $M$ and $p\in \psi(U)$.Let,the coordinates on $U$ be given by $(x_1,x_2,...,x_n)$.Now consider $\hat p\in U$ which maps to $p$.Then $d\psi_{\hat p}:T_{\hat p}U\to T_{p}M $ is a linear map which is the differential map and suppose $\partial^\varphi_i|_p=\frac{\partial^{\varphi}}{\partial x_i}|_p\in T_pM$ is defined as follows:
For $f\in C^\infty(M,\mathbb R)$, $\frac{\partial^\varphi }{\partial x_i}|_{p}(f):=\frac{\partial }{\partial x_i}|_{\hat p}(f\circ \varphi^{-1})$ where the partial derivative in the right hand side is the usual partial derivative w.r.t. $x_i$ at $\hat p$ of the function $f\circ \varphi^{-1}=f\circ \psi=:\hat f:U\to \mathbb R$ i.e. $\hat f\in C^\infty(U,\mathbb R)$.But,as such there is no inner product on $M$.So,if we want to use the same inner product(Obtained by pushing forward by $d\psi_{\hat p}$ that is on $T_{\hat p}( U)$,then $T_pM$ gets then same structural copy of the tangent space $T_{\hat p}U$ and in that case $d\psi_{\hat p}$ is an isometry between $T_{\hat p}U$ and $T_pM$ and if we call that inner product on $T_pM$ as $\langle.,.\rangle_p$,then $\langle \partial^{\phi}_i|_p,\partial^{\phi}_j|_p\rangle_p=\delta_{ij}$ i.e. they form an orthonormal basis on $T_pM$.But again,it is not something God given and is of little use.So,rather we can choose $\langle \partial^\varphi_i|_p,\partial^\varphi_j|_p\rangle=g_{ij}(p)$ such that $(g_{ij(p)})$ is symmetric and positive definite matrix of order $n\times n$.Now,if it happens so,that this inner product changes smoothly with the change of $p$ over the manifold $M$.Then we are calling it a Riemannian metric,am I right?
I mean what the value of $\langle\frac{\partial^\varphi}{\partial x_i}|_p,\frac{\partial^\varphi}{\partial x_j}|_p\rangle_p$ we are going to put is up to us,isn't it?They are not automatically orthonormal,as far as I think.
For example if $\psi:U=\mathbb R^2\to \mathbb R^2(=:M)$ is simply a linear map $\psi(x,y)=(x+y,y)$ then $e_1$ and $e_2$ are mapped to $e_1$ and $e_1+e_2$ respectively.But here $M=\mathbb R^2$ has a God given inner product coming from $\mathbb R^2$(that on the RHS) and so we have a natural choice to take $e_1,e_2$ as orthonormal and with respect to that inner product $e_1$ and $e_1+e_2$ are not orthogonal.But if we set $g_{ij}(p)=\langle \partial^\varphi_i|_p,\partial^\varphi_j|_p \rangle_p$,then we get that $e_1$ and $e_1+e_2$ are orthonormal.But the point is the God given inner product on $T_pM$ and the inner product copied from $T_{\hat p}U$ are different.In case of an abstract manifold with no preferred embedding in an Euclidean space,there is no such God given inner product to begin with,and we have to first define a metric on $T_pM$ which varies smoothly with $p\in M$.
I think what I am saying makes sense but I am not confident enough as I am a beginner.Can someone please clarify this point?