Good and bad tensors (and the metric) My question is the following : Why is the metric a good tensors that transforms "well".
A "good" tensor is a tensor that transforms like this :
I take a tensor $T^{\mu \nu}_\rho$
I thus have : $T=T^{\mu \nu}_\rho \partial_\mu \partial_\nu dx^{\rho}$
Now, if I change the coordinates, I will have : 
(*)
$$ \partial_\mu=\frac{\partial \widetilde{x}^{\alpha}}{\partial x^\mu}\widetilde{\partial}_{\alpha} $$
$$ dx^{\rho}=\frac{\partial x^{\rho}}{\partial \widetilde{x}^\mu}d\widetilde{x}^{\mu} $$
And the coordinates of my tensor in the new base $ \{ \widetilde{x} \}$will thus be :
$$ T^{\mu \nu}_\rho \frac{\partial \widetilde{x}^{\alpha}}{\partial x^\mu} \frac{\partial \widetilde{x}^{\beta}}{\partial x^\nu}\frac{\partial x^{\rho}}{\partial \widetilde{x}^\gamma}$$
We say that we have a "good" tensor if it transform well, it means that if I can write : 
$$\widetilde{T}^{\alpha \beta}_\gamma=T^{\mu \nu}_\rho \frac{\partial \widetilde{x}^{\alpha}}{\partial x^\mu} \frac{\partial \widetilde{x}^{\beta}}{\partial x^\nu}\frac{\partial x^{\rho}}{\partial \widetilde{x}^\gamma} $$
For example, $ (\partial_\mu \omega_\nu - \partial_\nu \omega_\mu) dx^{\mu}\wedge dx^{\nu}$ with $\omega$ a 1 form is a good tensor because if I start from  $ (\partial_\mu \omega_\nu - \partial_\nu \omega_\mu) dx^{\mu}\wedge dx^{\nu}$, and I use the transformations rules (*), I will end with an expression that will simply be $ (\widetilde{\partial}_\mu \widetilde{\omega}_\nu - \widetilde{\partial}_\nu \widetilde{\omega}_\mu) d\widetilde{x}^{\mu}\wedge d\widetilde{x}^{\nu}$.
On the opposite, $(\partial_\mu \omega_\nu + \partial_\nu \omega_\mu)dx^{\mu} dx^{\nu}$ is not a good tensor because if I start from this expression and I transform it using (*), I will end up with an expression that will not be of the form $(\widetilde{\partial}_\mu \widetilde{\omega}_\nu + \widetilde{\partial}_\nu \widetilde{\omega}_\mu)d\widetilde{x}^{\mu} d\widetilde{x}^{\nu}$.
Great.
To summarize : a "good" tensor has a mathematical expression that does'nt change if I write it on a map $ \{ \widetilde{x} \}$ or on a map $ \{ x \}$ : its expression doesn't depend on the map.
Now why is the metric tensor a "good" tensor.
Indeed, if I take for example the metric tensor associated to $ds^2=dx^2+dy^2$. I will have $g_{xx}=g_{yy}=1$ and the 2 others are $0$.
But if I work in polar coordinates, $g_{rr}=1$, $_{\theta \theta}=r^2$ and the 2 others are $0$.
So we see that I must define the metric tensor on a given map first and then I can deduce its expression on other maps. So in fact it is analog to $(\partial_\mu \omega_\nu + \partial_\nu \omega_\mu)dx^{\mu} dx^{\nu}$ : I can define it on a given map and then I deduce its expression on another map. But it is not possible to define the tensor unambigously without referring to any map.
So why do we say that the metric is a "nice" tensor that transform well ?
Is it because we define it as a scalar product. And a scalar product can be defined independently of any choice of basis ?
If possible I would like answers with things that I used in this post (im a big beginner in differential geometry).
 A: I think your question is caused by an unclear way to express things that unfortunately is quite common in differential geometry. The way out of the problem is partly indicated in your quesiton already. First of all, as indicated in the comment by @R.Alexandre , what you call a "bad tensor" does not really make sense as an object, so you should avoid the wording when thinking about the problem. 
The point is that tensors (or "good tensors" in the language you use) are geometric objects (like vector fields or one-forms). As you already indicate in the quesiton, the problem usually comes up when you try to define "operations" as the two operations you try to define on one-forms. So the "right" way to phrase the question is: "Given a geometric object of a certain type, does the following operation defined using local coordinates lead to a well defined tensor?". So you have to show that carrying out the same steps in a different local coordinate system leads to the same result (just expressed in different coordinates). If the operation does not involve differentiation, then this question boils down to linear algebra. For example the fact that a $\binom11$-tensor field has a well defined contraction which is a smooth function just means that the trace of a linear map from a vector space to itself does not depend on the basis used to obtain a matrix representation of the map. If the operation involves derivatives, then things are much more complicate, since then the freedom in the choice of local coordinates becomes hughe. (You can see this by  derivatives of chart changes entering the computations.) Consequently, there are only very few operations which a natural in that sense, like the Lie bracket of vector fields or the exterior derivative. There even are fairly complete classifications of these operations available. 
In that sense, you question about the metric is not completly easy to answer. In Riemannian geometry, the metric is defined to be a tensor. The context you mean probably is the one of hypersurfaces in Euclidean space. In this case, as you say, the metric is there as a geometric object (the restriction of the inner product on $\mathbb R^n$ to the tangent spaces of your hypersurface, which are linear subspaces in $\mathbb R^n$), and you just compute the local coordinate expressions of this geometric object.  
