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:
\begin{align}\large \color{red}{\mathbb{E}[{X^k}]} &= \displaystyle\int_{-\infty}^{\infty}\color{blue}{X^k}\,\,\color{green}{\text{pdf}}\,\,\,dx\tag{1}\end{align}
The moment generating function, $M_X(t)$, is a way to walk around this integral by, instead, carrying out:
\begin{align} \large\mathbb{E}[\color{blue}{e^{\,tX}}]&=\displaystyle \int_{-\infty}^{\infty}\color{blue}{e^{tX}}\,\color{green}{\text{pdf}}\, dx\tag{2}\end{align}
Why? Because it's easier and there is a fantastic property of the MGF that can be seen by expanding the Maclaurin series of $\color{blue}{e^{\,tX}}$ within the expectation operator:
$$M_X(t) = 1 + \frac{\color{red}{\mathbb{E} \left[X\right]}}{1!} \, t \, + \frac{\color{red}{\mathbb{E} \left[X^2\right]}}{2!} \, t^2 \, + \frac{\color{red}{\mathbb{E} \left[X^3\right]}}{3!} \, t^3 \, + \cdots\tag{3}$$
namely, 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 just once for all moments!
The fact that it is an easier integration is most apparent when the pdf is an exponential.
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.
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 train station (read, solve Eq $(2)$ just once), and from there access 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 $0$).