Let $f$, $g$ be functions of a real variable and let $F(f)$ and $F(g)$ be their fourier transforms. Then the fourier transform is linear in the sense that, for complex numbers $a$ and $b$,
$$F(af + bg) = a F(f) + b F(g)$$
i.e. it has the same notion of linearity that you may be used to from linear algebra. This is not a quirk - it expresses the fact that functions form an infinite dimensional vector space, with addition and multiplication by a scalar defined in the obvious way:
$$(f+g)(x) = f(x) + g(x)$$
$$(af)(x) = a f(x)$$
I'm glad you reposted this here; I saw it on MathOverflow earlier and wanted to answer, but it was closed (and it really is more appropriate here.)
As the answer there suggested, perhaps the best way to think about this is in terms of linear algebra.
Your sines and cosines form a complete orthonormal basis for your space of functions. The inner product on this space is $<f,g> = \int f(x)\bar{g(x)}dx$, and this gives a direct geometric interpretation of the integrals involved in calculating $a_n$ and $b_n$: you are simply projecting your function down onto a given basis vector to see "how much" of it is in "that direction." The $2/T$ factor out front comes from normalizing the basis vectors.
(For more details, see Bessel's inequality and Parseval's identity, or a good book like Stein and Shakarchi's 'Fourier Analysis: An introduction.' They have a wonderfully explicit visualization of this type on page 78, where they prove that partial sums $S_N(x)$ of Fourier series of a function f(x) are the best approximation possible with a trigonometric polynomial of order at most $N$.)
Your Fourier series, then, is equivalent to 'building up' f(x) from (an infinite linear combination of) basis vectors in the space, and the coefficients $a_n$ and $b_n$ tell you how much of each you need. The fact that you can do this at all is the content of Parseval's identity.
Best Answer
If you have a linear combination of functions, the resulting Fourier series is the corresponding linear combination of the Fourier series of the functions. So yes, it is linear. The key is that it is linear in the coefficients, even though the series is not linear in $x$.