On the meaning of s-plane and it's link to a transfer function
In the Laplace transform, being on the line $s = 0 \pm j\omega$ means you are looking at the pure sinusoids contained in the signal. This is exactly as you mentioned in your understanding of Fourier analysis. So this means that you are looking at the time-domain terms:

$$e^{j\omega{t}}$$

Which again represent the pure sinusoids composing your signal.

However once you are off this line and have some $s$ such that: $$s = \sigma \pm j\omega$$

Not only are you looking at the $e^{j\omega{t}}$ sinusoids terms, you are now looking at time domain terms:
$$e^{\sigma{t}}e^{j\omega{t}}$$

When talking about a stable system, poles being on the left-hand-plane of the $s$-domain means that $\sigma &lt; 0$ so the sinusoids you are considering contained in the signal now have a damping envelope. If $\sigma$ was positive, the poles would be placed in the right-hand-plane and the signal would explode to infinity as time went on. This is why we want poles to stay out of the right-hand-plane if we want a stable system.

The takeaway is that in Fourier analysis you are considering the pure sinusoids composing your signal. With the Laplace transform you are considering sinusoids with that exponential term attached. In the case for a stable or oscillating system, you are considering the damped sinusoids ($s = -\sigma \pm j\omega$), decaying exponentials ($s = -\sigma$), pure sinusoids ($s = \pm j\omega$), and constant amplitude ($s = 0$) components composing your signal.

This is what makes Fourier analysis appropriate for identifying frequency content and the Laplace transform appropriate for analyzing stability and performance parameters.