There is only one definition of "vector space" (up to isomorphism anyway), and that's what the author uses. You'll note that he doesn't talk about bases at all, the assumption of a basis is entirely in your mind. The entire point of the article is that the ℝ→ℝ function space is a vector space. A vector space is not required to have a basis, but assuming the axiom of choice, every vector space does have (at least) one, including that of ℝ→ℝ functions.

You are misreading the article. At a guess, you have come into it "knowing what a vector is" from physics or computer science. Specifically, a vector is not "a sequence of reals", it is any object that obeys a certain set of properties. He is not "extending" your definition, he's trying to teach you the actual definition, which is more general than the examples you are used to.

(Imagine/remember what it feels like when children first learn that the integers aren't the only number, there are also fractions, then irrationals, then complex numbers...this is a very similar situation).

With that in mind, you may want to reread the text and pay attention to the definitions he is using, and not assume that your definitions are the whole story.

Sorry, I couldn't find the page in English, but what you're talking about is a Hilbert basis: https://fr.wikipedia.org/wiki/Base_de_Hilbert . There is a paragraph on this in the orthonormal basis English page: https://en.wikipedia.org/wiki/Orthonormal_basis

Another example is the eigenvectors of linear operators like the Laplacian. Recall how, in finite dimension, the eigenvectors of a full rank operator (matrix) form an orthonormal basis of the vector space. There is a similar notion in infinite dimension. I can't find an English page that covers this very well, but there's a couple of paragraphs in the Spectral Theorem page (https://en.wikipedia.org/wiki/Spectral_theorem#Unbounded_sel... ). The article linked here also touches on this.

Regarding your last sentence, one thing to note is that having a basis is not what makes you a Hilbert space, but rather having an inner product! In fact, to get the Fourier coefficients, you need to use that inner product.

No, and this is where this formal notion of basis I mentioned unfortunately diverges from what is perhaps more useful in practice.

You can represent any function f: [-pi, pi] -> R as an infinite sum

    f(x) = sum_(k = 0 to infinity) (a_k sin(kx) + b_k cos(kx))

for some coefficients a_k and b_k as long as f is sufficiently nice (I don't remember the exact condition, sorry).

This is very useful, but the functions sin(x), sin(2x), ... , cos(x), cos(2x), ... don't constitute a basis in the formal sense I mentioned above as you need an infinite sum to represent most functions. It is still often called a basis though.

You can’t, because a Fourier series is not a linear combination.

If you're familiar with Zorn's Lemma, the construction is just to order bases by inclusion and to consider chains created by noting that there must be an independent dimension and adding it inductively. You can upper bound each of these chains by unioning the members of the chain (which preserves linear independence). By Zorn's Lemma that means there is a maximal linearly independent system and if an element existed outside of that system's span it would contradict that maximality.

He doesn't need to talk about it (though you might like to look up the notorious Hilbert's Basis Theorem); it happens to be the case that any vector space has a basis, but even if you don't know that, a vector space is still a vector space and its elements are still vectors.