I could give an example of what people mean when they "say":
... metric tensor depend on the local coordinate system and therefore are not intrinsic to the surface
Take for example the Schwarzschild metric. We have
$$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2 + \left(1-\frac{2m}{r}\right)^{-1}dr^2 +r^2(d\theta^2 +\sin^2\theta d\phi^2) $$
If you read this as $$ds^2 = dx_{\mu} \ g^{\mu \nu} \ dx_{\nu}$$
there are many choices available to you. Three (easier) examples that serve to clarify the point could be:
- Pick your co-ordinates to be $dt, dr, rd\theta, r\sin \theta d\phi$, and then your metric tensor reads ${\rm diag}(-(1-2M/r), (1-2M/r)^{-1}, 1,1)$.
- Co-ordinates are $dt, dr, d\theta, d\phi$, and your metric tensor reads ${\rm diag}(-(1-2M/r), r^2(1-2M/r)^{-1}, r^2,r^2 \sin^2 \theta)$.
- A hopeless choice - Deliberately ensure that you have a Minowski metric ${\rm diag}(-1,1,1,1)$ and mess up your "co-ordinates" (absorbing square root of that (1-2M/r) factor in your co-ordinate definitions).
Of course, there are many other choices. But this conveys the basic premise. Metric tensor does depend on what co-ordinate system you are adopting.
Also, while adopting co-ordinates that are easier to work with is understandable, changing your co-ordinates isn't going to make a difference to the basic physics associated with it. For example, in the above case, you can't avoid the fact that EITHER your co-ordinates (with all the redefinition possibilities), OR your metric tensor, will become singular at $r = 2M$, with everything it leads to! (Also at $r=0$ but that's not as glaring, to me at least). One can ease the maths, but can't change the Physics by changing co-ordinates!