Why the surface of the sphere is not a Euclidean space?

Possibly what's confusing you is that the statement refers to the surface of the sphere, not the space the sphere is sitting in. If we ignore relativistic effects (and the fact that the Earth isn't quite a sphere), yes the sphere is in a Euclidean 3-dimensional space. In that space all the convenient Euclidean things apply. But the statement didn't refer to the sphere, or the space it's sitting in: it referred to the surface of the sphere. The surface of the sphere is a 2-dimensional space, not a 3-dimensional one, and points on it only need 2 coordinates (latitude and longitude, for example).

To help see the difference, consider a straight line between London and New York. In the 3-dimensional Euclidean space in which the Earth is embedded, that straight line goes through the Earth. But if we're only considering the surface of the Earth, that line doesn't exist. The straight line (shortest distance between the two points) on the surface lies along the great circle. Now consider drawing the lines from both New York and London to, say, Capetown, to make a triangle. Yes, if you draw the lines through the Earth you will get a nice Euclidean triangle with angles that add up to 180 degrees. But those lines don't exist in the space you are considering: you can only draw lines on the surface of the Earth. The angles of the triangle drawn on the surface of the Earth add up to more than 180 degrees, so the space must be non-Euclidean.

Edit: The bit after the "but" just seems to be saying that for most purposes you can treat a smallish bit of the Earth as if it were flat. You probably don't need to worry about the curvature of the Earth when looking at a street-map of your town.

Intuitively, Euclidean geometry is flat. The sphere, globally, is clearly not flat. But locally, if you tear out a tiny piece of it it's pretty much flat.

More accurately, in a Euclidean space Euclid's fifth postulate holds. Thus, given a line and a point not on that line there exists a unique parallel to the line through that point. This fails on the sphere since there are no parallel lines at all. Every two straight lines (where straight should be interpreted as geodesics, that is straight relative to the curved geometry of the sphere, thus straight lines are in correspondence with great circles) intersect. However, the geometry of a small piece of the sphere is locally Euclidean, that is it looks and behaves just like a little piece of $\mathbb R^2$.

(1) Topologically, Euclidean space is retractable to a point. That means there is a continuous map: $f:\mathbb R^n\times I\to \mathbb R^n$ such that $f(\textbf v,0)=\textbf v$ and $f(\textbf v,1)=\textbf 0$.

A sphere is not retractible to a point. There is no similar map $f:S^{n}\times I\to S^{n}$ (not a trivial result.) So $S^{n}$ is not topologically the same as any $\mathbb R^{m}$.

(2) Any point on $S^{n}$ has a neighborhood that is homeomorphic to some open subset of $\mathbb R^n$. Such a homeomorphism from a subset of $\mathbb R^n$ to an open subset of a space $M$ is often called a "chart" because it gives a coordinate system in that open subset.