As I am a beginner in differential geometry, my question may be ill defined because everything is not clear for me yet, but I think it is ok.
When we do differential geometry, we can define vectors without refering to a given basis. To do it, we use curves $\gamma : \mathbb{R} \rightarrow M$ (M is the manifold) and we say that the vector is the derivative of $\gamma(t)$ at $t=0$.
Then we write it in a basis and we have the expression of the vector in any coordinates we want (polar, cartesian and anything else).
But for the metric I haven't found if we can do the same.
Indeed if we use the "classical" metric $ds^2=dx^2+dy^2+dz^2$, we write it in the cartesian basis (as I just did) and then by change of coordinates we are able to find it in other basis.
But if I understood well, the global philosophy of differential geometry is precisely to work with quantities that are independant of the set of coordinates we choose. This is why we define tensors as tensorial products of vectors and covectors. By doing so we ensure the fact that the expression of tensors will have the same form in all system of coordinates (correct me if I misunderstood this "philosophy" of tensors please, or say me if it is ok, that's a part of my question !).
For the metric, I know that we say $g_{ij}=<\partial_i, \partial_j>$ and this form is independant of the system of coordinates chosen, but in fact, when we say that we say nothing as we don't know the result of such a calculation.
So : Is it possible to define the metric without refering to any system of coordinates like we did for vectors with the curves ? Because if it is not possible the fact that we say the metric is independent of the system of coordinates chosen seems a little wrong to me ? Indeed we could never guess its expression without doing a change of coordinates from the basis it has been defined.
The definition of the riemannian metric is independent of a choice of coordinates, because a riemannian metric is a smooth (I don't want to go into detail here) map $g$, which associates to every point $p\in M$ a inner product $g_p$ on the tangent space $T_pM.$ So you see that we start out with manifold and put a metric on it and that this does not depend on any coordinate chart.
The maps $g_{ij}(p):=g_p(\partial_i,\partial_j)$ you were reffering to are called the local representation of the metric and these certainly depend on your chosen chart!