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Let's assume that an equation-free intuition is not possible, and still insist on boiling down the math to the very essentials to get an idea of what's going on: we are trying to obtain the statistical moments, which, after the obligatory reference to physics, we define as the expected value of a power of a random variable. For a continuous random variable, the raw $k$-th moment is by LOTUS:

the moments appear "perched" on this polynomial "clothesline", ready to be culled by simply differentiating $k$ times and evaluating at zero once we go through the easier integration (in eq. (2)) just once for all moments! The fact that it is an easier integration is most apparent when the pdf is an exponential.

Let's assume that an equation-free intuition is not possible, and still insist on boiling down the math to the very essentials to get an idea of what's going on: we are trying to obtain the statistical raw moments, which, after the obligatory reference to physics, we define as the expected value of a power of a random variable. For a continuous random variable, the raw $k$-th moment is by LOTUS:

the raw moments appear "perched" on this polynomial "clothesline", ready to be culled by simply differentiating $k$ times and evaluating at zero once we go through the easier integration (in eq. (2)) just once for all moments! The fact that it is an easier integration is most apparent when the pdf is an exponential.

In response to the question under comments about the switching from $X^k$ to $e^{tx}$, this is a completely strategic move: one expression does not follow from the other. Here is an analogy: We have a car of our own and we are free to drive into the city every time we need to take care of some business (read, integrating Eq $(1)$ no matter how tough for every separate, single moment). Instead, we can do something completely different: we can drive to the nearest subway station (read, solve Eq $(2)$ just once), and from there use public transportation to reach every single place we need to visit (read, get any $k$ derivative of the integral in Eq $(2)$ to extract whichever $k$-th moment we need, knowing (thanks to Eq $(3)$) that all the moments are "hiding" in there and isolated by evaluating at $0$).

In response to the question under comments about the switching from $X^k$ to $e^{tx}$, this is a completely strategic move: one expression does not follow from the other. Here is an analogy: We have a car of our own and we are free to drive into the city every time we need to take care of some business (read, integrating Eq $(1)$ no matter how tough for every separate, single moment). Instead, we can do something completely different: we can drive to the nearest subway station (read, solve Eq $(2)$ just once), and from there use public transportation to reach every single place we need to visit (read, get any $k$ derivative of the integral in Eq $(2)$ to extract whichever $k$-th moment we need, knowing (thanks to Eq $(3)$) that all the moments are "hiding" in there and isolated by differentiating and evaluating at $0$.

The moment generating function, $M_X(t)$, is a way to walk around this integral by, instead, carrying out:

The fact that eventually there is a need to differentiate makes it a not a free lunch - in the end it is a Laplacian transform.

So we may be allowed to imagine that the $M_X(t)$ function decomposes the $\text{pdf}$ somehow into its "constituent frequencies" as spelled out in Eq $(3)$?

The moment generating function, $$M_X(t):=\mathbb E\big[e^{tX}\big],$$ is a way to walk around this integral (Eq.1) by, instead, carrying out:

The fact that eventually there is a need to differentiate makes it a not a free lunch - in the end it is a two-sided Laplace transform of the pdf with a changed sign in the exponent:

$$\mathcal L \{\text{pdf}(x)\}(s) =\int_{-\infty}^{\infty}e^{-sx}\text{pdf}(x) dx$$

such that $$M_X(t)=\mathcal L\{\text{pdf}(x)\}(-s)\tag 4.$$

Therefore the $M_X(t)$ function decomposes the $\text{pdf}$ somehow into its "constituent frequencies" when $\sigma=0.$ From eq. (4):

\begin{align}\require{cancel} M_X(t)&=\mathbb E\big[e^{-sX}\big]\\[2ex] &=\displaystyle \int_{-\infty}^{\infty}{e^{-sx}}\,\text{pdf}(x)\, dx\\[2ex] &=\displaystyle \int_{-\infty}^{\infty}{e^{-(\sigma+i\omega)x}}\,\text{pdf}(x)\, dx\\[2ex] &=\displaystyle \int_{-\infty}^{\infty}\cancel{e^{-\sigma x}}\,\color{red}{e^{-i\omega x}\,\text{pdf}(x)\, dx} \end{align}

which leaves us with the improper integral of the part of the expression in red, corresponding to the Fourier transform of the pdf.

In general, the intuition of the Laplace transform poles of a function would be that they provide information of the exponential (decay) and frequency components of the function (in this case, the pdf).