How is the Fourier transform "linear"?

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:

$$(f+g)(x) = f(x) + g(x)$$ $$(af)(x) = a f(x)$$

Linear in this context has nothing to do with the geometry of the graphs of functions under consideration. It is used in the sense that the functions themselves form a vector space, i.e., you can add functions and take multiples of a function with a real number. The linearity of the fourier transform means that if you take the transform of a sum of functions, it is the same as the sum of the fourier transforms of the functions, and the same holds for real multiples of functions.

The Fourier transform is linear as a function whose domain consists of functions, that is, the sum of the Fourier transforms of two functions is the same as the Fourier transform of the sum. Same with scalars. For more information, see Properties of the Fourier transform (Wikipedia).

The term linear is actually fairly consistently used. That is, a transformation $T$ of vector spaces is called linear if $T(ax+by)=aT(x)+bT(y)$ for scalars $a,b$. If you think of things this way, you will see that your favourite linear functions on $\mathbb{R}$ are specific cases of this.