Consider two periodic functions. Assume their sum is not periodic. The periodic functions can be represented by a Fourier series. If you add up the Fourier series, you get a series that represents their sum. But their sum is not periodic, yet you have described it using a Fourier series.
I thought that non-periodic functions can't be represented by a Fourier series. Why isn't this a contradiction?
On
The Fourier series of a non-periodic function is really the Fourier series of its periodic extension. For example, there is a Fourier series of $f(x)=x$ on $[0,\pi]$, which is actually the Fourier series of the sawtooth wave that is formed by $\pi$-periodically extending $f(x)=x$. The Fourier series for a non-periodic function will not converge at every point but will still converge in the sense of $L^2$.
Also, Fourier series are not meant to be defined on the whole line, they are indeed meant to be defined on intervals. This has to do with the changes in the spectrum of the Laplacian as the domain increases: in the limit, the spectrum becomes dense, and you have to turn to the Fourier transform instead of Fourier series.
On
A Fourier series means the amplitude of the different harmonics, who are an integer multiple of a base frequency.
It is easy to see, that this base frequency simply doesn't exist in your case.
Although a Fourier transform of a such function of course exist, which is trivially
$$F(s)=\delta(t-\omega_1)+\delta(t-\omega_2)$$
In order to find its Fourier series, a periodic function with period $R$ should be thought of as a function defined on a circle of circumference $R$, call it $S^1_R$. The Fourier series of the function is then its representation in the basis of $L^2(S^1_R)$ given by orthonormal eigenfunctions of the Laplace operator.
If two functions have incommensurate periods, then their sum is nonperiodic, does not descend to a circle of any circumference, and therefore does not have a Fourier series.
As functions on $\mathbb{R}$, if they are sufficiently nice, the Fourier transform gives an analogous decomposition, but because there are so many more eigenfunctions of the Laplace operator on $\mathbb{R}$, the sum is an integral. Compare:
Let $e_\omega(t) = e^{2\pi i\omega t}$ for $\omega$ real. $$ \mbox{periodic ($\omega$ is an integral multiple of a base frequency): }\\ f(t) = \sum_\omega\langle f,e_\omega(t)\rangle e_\omega$$ $$ \mbox{nonperiodic ($\omega$ ranges over $\mathbb{R}$): }\\ f(t) = \int\langle f,e_\omega(t)\rangle e_\omega\ d\omega$$ where $$\langle f,e_\omega\rangle = \int f(x)\overline{e_\omega(x)}\ dx$$ You will recognize that if we integrate over the circle, $\langle f,e_\omega\rangle$ gives the series of Fourier coefficients and if we integrate over $\mathbb{R}$, it is the Fourier transform.