The â€śgut intuitionâ€ť youâ€™re looking for is the notion of a *projection.*

It will help if you draw pictures while going through the following explanation:

Imagine two-dimensional Euclidean space, \mathbb{R}^2 (i.e. the standard x-y plane). Ordinarily, we notate points in this plane by their components in the set of basis vectors \{i, j\}, where i is the unit vector in the +x direction, and j is the unit vector in the +y direction.

How do we find the coordinates of a given vector v in this basis? Well, so long as the basis is *orthogonal* (i.e., the basis vectors are at right angles to each other), we simply take the *orthogonal projection* of v onto i and j. Intuitively, this means finding â€śthe amount of v that points in the direction of i or j.â€ť More formally, we can calculate it with the dot-product - the projection of v onto any other vector w is equal to \frac{v \cdot w}{|w|}. (Since i and j are *unit vectors* we see simply that the coordinates of v are v \cdot i and v \cdot j.) (We can also see that â€śorthogonalâ€ť can be defined as â€śhas zero dot product.â€ť)

But we can do this same process to find the coordinates of v in *any* orthogonal basis. For example, imagine the basis \{i+j, i-j\} - the coordinates in this basis are given by \frac{v \cdot (i + j)}{\sqrt{2}} and \frac{v \cdot (i - j)}{\sqrt{2}}. Let us now â€śunwrapâ€ť the formula for dot product, and look a bit more closely:

\frac{v \cdot (i + j)}{\sqrt{2}} = \frac{1}{\sqrt{2}}\sum_{n}{v_{n}(i+j)_{n}}

So, what have we really done to change coordinates? We expanded both v and i+j in a basis, multiplied their components, and added them up.

Now, the previous example was only a change of coordinates in a finite-dimensional vector space.

However, as we will see, the core idea does not change much when we move to more-complicated structures. Observe the formula for the Fourier transform:

\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)\ e^{-2\pi i x \xi}\,dx, where \xi \in \mathbb{R}

What we need to see, here, is that this is fundamentally the same formula that we had before. f(x) has taken the place of v_{n}, e^{-2\pi i x \xi} has taken the place of (i+j)_{n}, and the sum over n has turned into an integral over dx, but the underlying concept is precisely the same. To change coordinates in a *function space*, we simply take the orthogonal projection onto our new basis *functions*. In the case of the Fourier transform, the function basis is the family of functions of the form f(x) = e^{-2\pi i x \xi} for \xi \in \mathbb{R}. Since these functions are oscillatory at a frequency determined by \xi, we can think of this as a â€śfrequency basis.â€ť

Now, the Laplace transform is somewhat more complicated - as it turns out, the Fourier basis is orthogonal, and so the analogy to the simpler vector space holds almost-precisely. The Laplace basis is *not* orthogonal, and so we canâ€™t interpret it *strictly* as a change of coordinates in the traditional sense. However, the intuition is precisely the same: we are taking the orthogonal projection of our original function onto the functions of our new basis set:

F(s) =\int_0^\infty f(t)e^{-st} \, dt, where s \in \mathbb{C}

Here, it becomes obvious that the Laplace transform is a *generalization* of the Fourier transform, in that the basis family is strictly larger (we have allowed the â€śfrequencyâ€ť parameter to take *complex* values, as opposed to merely *real* values). The upshot of this is that the Laplace basis contains functions that grow and decay, while the Fourier basis does not.

I hope this clears it up a bit.