What is the intuition behind Gauss sums?

When you say $\chi$ is a character of the field $F_p$, you really mean it is a character of the group $F_p^\times$. Any (multiplicative) character $\chi$ on $F_p^\times$ can be extended to a function on $F_p$ by setting $\chi(0) = 0$, and with this convention $\chi$ as a function on $F_p$ is totally multiplicative. Fixing a choice of nontrivial $p$th root of unity $\zeta$, any function $f \colon F_p \rightarrow {\mathbf C}$ has a Fourier transform ${\mathcal F}f \colon F \rightarrow {\mathbf C}$ given by $({\mathcal F}f)(a) = \sum_{t \in F_p} f(t)\overline{\zeta^{at}}$. So the Gauss sum $g_a(\chi)$ is essentially the Fourier transform at $a$ of the function $\chi$ (as a function on $F$). For more on Fourier transforms of functions on a finite abelian group, see Section 4 (starting at Definition 4.4) of
https://kconrad.math.uconn.edu/blurbs/grouptheory/charthy.pdf

Another intuition (besides the idea that a Gauss sum of a character is basically the Fourier transform of that character, viewed as a function on the additive group $F_p$) is that a Gauss sum is a discrete analogue of the Gamma function. See pp. 56--58 of Koblitz's book "$p$-adic Analysis: A Short Course on Recent Work" for a table illustrating this analogy (including the idea that a Jacobi sum is like the Beta function).

One explanation is given in this math.SE answer. In the language of that answer, you want to describe the unique quadratic subfield of $\mathbb{Q}(\zeta_p)$. Since it's quadratic, it's generated by the square root of some rational, so the Galois group acts by multiplication by $-1$ on it. Therefore you want to find an element of $\mathbb{Q}(\zeta_p)$ such that the Galois group $(\mathbb{Z}/p\mathbb{Z})^{\ast}$ acts by multiplication by $-1$ on it, and up to a constant that element is a Gauss sum (for the quadratic character). Gauss sums with respect to more general characters have a similar relationship to actions of the Galois group.

They are the discrete Fourier transform of the character $\chi$ (which presumably is a multiplicative character -- like the quadratic symbol, no?). This allows one to express $\chi$ as a sum of exponentials.