Hi,
Excuse me for carelessly throwing weird terms around. Since you were considering Riemannian manifolds I figured you would be familiar with "smooth structure".
Note that in the definition of Taylor expansion on Euclidean space you use that the function is differentiable (as often as you need) at the point where you take the Taylor expansion. So simply asking the function to be continuous will in general not be enough. For instance the derivative of the function x
⅓ is ill-defined at x=0, although this function is definitely continuous at x=0.
So in general when we consider Taylor expansions we consider these infinitely differentiable functions, instead of "infinitely differentiable" people often say "smooth". There are notions around for less differentiable functions, but they are not widely used and definitely not standard (by which I mean I don't know enough of them to tell you anything coherent).
In calculus (as developed by Newton or Leibniz or whatever other smart fellow) we learn how to differentiate functions f:ℝ
n → ℝ. A Riemannian manifold is first of all a "smooth" manifold. This means that we can find coordinates around every point such that the coordinate transformations on overlaps of these coordinates are smooth, i.e. infinitely differentiable. This allows us to define differentiation, and thus smoothness, of functions on the manifold. We simply do it around each point as we do it in calculus.
https://en.wikipedia.org/wiki/Differentiable_manifold
There are many excellent books on this topic. I used the book introduction to manifold by Tu.
The Riemann normal coordinates are a special kind of local coordinates that respect certain properties of the Riemannian metric on the Riemannian manifold. So you could indeed use these. My point above was however that the Taylor expansion of a function (which should then be smooth) will not depend on the Riemannian metric. You could calculate it in any local coordinate system around the point.
Now about the simple function x
2. Let's say we have a smooth function on our manifold and around the point x
0 we can "center" a coordinate system such that this function looks like x
2. By center I mean we have a coordinate system such that x
0=0 in those coordinates. The the Taylor expansion is exactly x
2. both the function and it's first derivative vanish at x
0, while ½ times the second derivative is 1 at x
0 and the higher derivatives also vanish. If we pick some coordinate system around x
0 such that the function looks like x
2 (but not a centered one) then we get the Taylor expansion x_0^2+ 2x_0(x-x_0) +(x-x_0)^2. Note that we could then have picked the coordinate system y=x-x_0 instead which is centered around x_0 only then the function looks like (y+x_0)^2. We could then Taylor expand with respect to y instead.
I hope this helps a bit, in the last paragraph I was a bit unclear since I used x_0 both to denote the point and the value of the point in the coordinate system x. If anything remains unclear or you would like some references to good sources on this topic let me know!
(sorry for the bad typesetting I only figured out how to tex on this forum again after a while).