Starting from $s_0=2$ and the easily computable data (thanks to Ted) $\#E(F_5)=6=5+1-t$ that tells us $t=s_1=0$, we get $s_2=ts_1-5s_0=-10$. Therefore the formula tells us that $\#E(F_{25})=5^2+1-(-10)=36.$
In order to clear any potential misunderstandings I will list all the 36 points. To that end we first need to construct the field $F_{25}$. I constructed it using the primitive polynomial $x^2+3x+3$, so I use $F_{25}=F_5[x]/\langle x^2+3x+3\rangle$. I will denote the coset $x+\langle x^2+3x+3\rangle$ by $\alpha$, so the arithmetic will follow from the rule $\alpha^2=2+2\alpha$.
Let us list the discrete logarithms (base $\alpha$) of the elements of $F_{25}^*$: $ \begin{array}{l|ll|ll|l} n&\alpha^n&&&n&\alpha^n\\ \hline 0&1&&&12&4\\ 1&\alpha&&&13&4\alpha\\ 2&2+2\alpha&&&14&3+3\alpha\\ 3&4+\alpha&&&15&1+4\alpha\\ 4&2+\alpha&&&16&3+4\alpha\\ 5&2+4\alpha&&&17&3+\alpha\\ 6&3&&&18&2\\ 7&3\alpha&&&19&2\alpha\\ 8&1+\alpha&&&20&4+4\alpha\\ 9&2+3\alpha&&&21&3+2\alpha\\ 10&1+3\alpha&&&22&4+2\alpha\\ 11&1+2\alpha&&&23&4+3\alpha\\ &&&&\infty&0 \end{array} $ Now it becomes easy to list the points in $E(F_{25})$. As we are in characteristic five, we might as well use the equation $y^2=x^3+2$. For a given $x$ we have that $y^2=x^3+2$ has one solution, iff $x^3+2=0\Leftrightarrow \log (x^3+2)=\infty$; two solutions, iff $\log(x^3+2)$ is even; and no solutions, iff $\log(x^3+2)$ is odd. Furthermore, $\alpha^8$ is a primitive cubic root of unity, so $\log((\alpha^n)^3+2)$ is periodic with period 8. We get three full periods while $x$ runs over the elements of $F_{25}^*$.
Using the above table we see that when $\log x$ takes the respective values $0,1,2,3,4,5,6,7$, then $\log(x^3+2)$ takes the respective values $6,8,\infty,23,0,16,12,19$. There are 5 even values in the latter list and a single $\infty$, so in the range $0\le\log x<8$ we get 11 points on the curve. With $x$ ranging over all of $F_{25}^*$ we get thus $3\cdot11=33$ points. Add the point at infinity as well as the points $x=0$, $y=\pm\sqrt{2}=\pm(2+3\alpha)$, and we get altogether $ 33+1+2=36\ \text{points}\ \in E(F_{25}). $
For the sake of completeness I list the points. Let $\omega=\alpha^8=1+\alpha$ be a fixed primitive cubic root of unity. Then we have
- 6 points of the form $(\omega^j,\pm\alpha^3)$, $j=0,1,2$,
- 6 points of the form $(\omega^j\alpha,\pm\alpha^4)$, $j=0,1,2$,
- 3 points of the form $(\omega^j\alpha^2,0)$, $j=0,1,2$,
- 6 points of the form $(\omega^j\alpha^4,\pm1)$, $j=0,1,2$,
- 6 points of the form $(\omega^j\alpha^5,\pm\alpha^8)$, $j=0,1,2$,
- 6 points of the form $(3\omega^j,\pm2)$, $j=0,1,2$,
- 2 points $(0,\pm\alpha^9)$,
- and the point at infinity.
Remembering that $\alpha^k\in F_5$, iff $6\mid k$ allows us to identify the six points in $E(F_5)$ on this list.