Tate twists have to do with fundamental classes in etale (co)homology.
Usually we talk about etale cohomology, but we can also talk about homology,
just as the dual of cohomology.
If $X$ is smooth and projective of dimension $d$, then $H^{2d}(X)$ will be one-dimensional, and transform under $\chi^{-d}$; equivalently, $H_{2d}(X)$ will be one-dimensional, and transform under $\chi^d$.
How do you see this? Of course there is a rigorous proof that you can read, but intuitively, if you think about how fundamental classes are constructed in topology, you can convince yourself that a fundamental class in dimension $d$ will behave like a $d$-fold product of a fundamental class in dimension $1$.
So we reduce to the one-dimensional case.
It is then easy to see that there is a canonical isomorphism
$$H_2(\mathbb P^1) \cong H_1(\mathbb A^1\setminus \{0\}).$$
Finally, this last homology is the same as the $\ell$-adic Tate module
of $\mathbb A^1 \setminus \{0\}$ (thought of as the multiplicative group),
and that Tate module is one-dimensional, with Galois action given by $\chi$.
As for the Tate conjecture:
Imagine that $Y$ is a smooth connected closed subvariety of codimension $j$ of $X$, itself smooth and projective, say, of dimension $d$.
Let's work with homology first, because it's more intuitive.
The map $Y\to X$ will induce
a map $H_{2(d-j)}(Y) \to H_{2(d-j)}(X)$. Now since $Y$ has dimension $d-j$, the source of this map is one-dimensional, with
Galois action given by $\chi^{d-j}$. So a codimension $j$ cycle
gives a Galois-invariant line in $H_{2(d-j)}(X)$ (the image of the above map) which transforms via $\chi^{d-j}$.
Remember that I said $H_{2(d-j)}(X)$ was just the dual of $H^{2(d-j)}(X)$.
Now cup product gives a perfect pairing (by Poincare duality)
$$H^{2(d-j)}(X) \times H^{2j}(X) \to H^{2d}(X),$$
compatible with Galois. Since $X$ is $d$-dimensional, the Galois action
on the one-dimensional space $H^{2d}(X)$ is via $\chi^{-d}$.
So we see that as Galois reps., the dual to $H^{2(d-j)}(X)$ can be identified
with $H^{2j}(X)\otimes \chi^d$.
Thus our codimension $j$ cycle gives a line in $H^{2j}(X)\otimes \chi^d$ which transforms under Galois via $\chi^{d-j}$, which is the same
thing as an invariant line in $H^{2j}(X)\otimes \chi^j$.
So codimension $j$ cycles contribute invariant lines in $H^{2j}(X)(j)$,
and the Tate conjecture is that the full space of Galois invariants here is actually spanned by the lines coming from cycles.
I figured I'd come back to this and expand my comment into an answer.
Everything you've done so far looks good. You just need to prove that
$$\sum_{i=2}^{p-1} \chi(i) = -1,$$
or, equivalently,
$$\sum_{i=1}^{p-1} \chi(i) = 0.$$
To prove this last equality, note that the set of integers from $1$ to $p-1$ is a set of representatives for $(\Bbb{Z}/p\Bbb{Z})^\times$. Thus if $a\in(\Bbb{Z}/p\Bbb{Z})^\times$, then $\{ai : 1\le i \le p-1\}=\{i : 1\le i \le p-1\}$, so
$$\chi(a)\sum_{i=1}^{p-1} \chi(i)
= \sum_{i=1}^{p-1} \chi(a)\chi(i)
= \sum_{i=1}^{p-1} \chi(ai)
= \sum_{i=1}^{p-1} \chi(i),
$$
and since $\chi$ is nontrivial, there is some $a\in (\Bbb{Z}/p\Bbb{Z})^\times$ such that $\chi(a)\ne 1$. Thus, this equality tells us that when $\chi$ is a nontrivial character,
$$\sum_{i=1}^{p-1} \chi(i)=0,$$
as desired.
Best Answer
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).