$S(u)\equiv 2p\pmod u $? shown that
Let $\sum_{q=1}^{u}q^{u-1}= S(u)$

if $u\in \{3p\}$ here $p$ be a odd prime, then $S(u)\equiv 2p\pmod u $

Simply $S(3p)\equiv 2p\pmod {3p} $
Example
Let $p=5\rightarrow u=15 \rightarrow S(15)\equiv 10\pmod {15} $
Source code
forprime(p=3, 100, print ([3*p,sum(q=0,3*p,q^(3*p-1))%(3*p)]))
[9, 6]
[15, 10]
[21, 14]
[33, 22]
[39, 26]
[51, 34]
[57, 38]
[69, 46]
[87, 58]
[93, 62]
[111, 74]
[123, 82]
[129, 86]
[141, 94]
[159, 106]
[177, 118]
[183, 122]
[201, 134]
[213, 142]
[219, 146]
[237, 158]
[249, 166]
[267, 178]
[291, 194]

I observe this property not found for other prime except $3$. Thank you.
 A: By Fermat's little theorem, for all $q$ which are not a multiple of $p$, you have
$$q^{p-1} \equiv 1 \pmod p \implies q^{3(p-1)} \equiv 1 \pmod p \implies q^{3p-1} \equiv q^2 \pmod p \tag{1}\label{eq1A}$$
Note the above is also true where $q$ is a multiple of $p$. Thus, you have
$$\begin{equation}\begin{aligned}
S(u) & = \sum_{q=1}^{u}q^{u-1} \\
& \equiv \sum_{q=1}^{u}q^2 \pmod p \\
& \equiv \frac{u(u + 1)(2u+1)}{6} \pmod p \\
& \equiv \frac{3p(3p + 1)(6p+1)}{6} \pmod p \\
& \equiv \frac{p(3p + 1)(6p+1)}{2} \pmod p \\
& \equiv 0 \pmod p
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note I used the formula for the sum of consecutive squares given in Induction and the sum of consecutive squares. Thus, modulo $u = 3p$, the result would be $0$, $p$ or $2p$.
Note that since $u - 1$ is even, you have $q^u \equiv 0 \pmod 3$ if $q \equiv 0 \pmod 3$ and $q^u \equiv 1 \pmod 3$ otherwise. Thus, for each group of $3$ items being added, they have a sum of $2$, so for up to $u = 3p$, the overall sum would be $2p$. Thus, you have
$$S(u) \equiv 2p \pmod 3 \tag{3}\label{eq3A}$$
You've already shown the result is $2p$ where $p = 3$. For $p \neq 3$, since $p$ and $3$ are then relatively prime, the Chinese remainder theorem shows you can combine \eqref{eq2A} with \eqref{eq3A} to get a unique solution, with this being
$$S(u) \equiv 2p \pmod u \tag{4}\label{eq4A}$$
as requested.
