Here I'll give the coset representatives that are given in Lemma 5.1.1 of RobertsSchmidt. Note that they use the symplectic form $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}$ as well as the transpose on the first factor. Therefore, we have to change bases to get the correct elements for our group. Recall this amounts to conjugating by $P = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}$.
(x, y, z) (x, y, z) 
[1 0 0 0] [0 1 0 0] [0 0 0 1] [0 0 1 0] [1 0 0 0] [0 1 0 0] [0 0 0 1] [0 0 1 0] 
They give elements $s_1$ and $s_2$ that arise from the Bruhat decomposition. We change those to $S_1$ and $S_2$ for us. In fact, $S_1 = s_1$, but $S_2$ is not the same as $s_2$.
[0 1 0 0] [1 0 0 0] [0 0 0 1] [0 0 1 0] [0 1 0 0] [1 0 0 0] [0 0 0 1] [0 0 1 0] 
[ 1 0 0 0] [ 0 0 1 0] [ 0 1 0 0] [ 0 0 0 1] [ 1 0 0 0] [ 0 0 1 0] [ 0 1 0 0] [ 0 0 0 1] 
[0 1 0 0] [1 0 0 0] [0 0 0 1] [0 0 1 0] [0 1 0 0] [1 0 0 0] [0 0 0 1] [0 0 1 0] 
[ 1 0 0 0] [ 0 0 0 1] [ 0 0 1 0] [ 0 1 0 0] [ 1 0 0 0] [ 0 0 0 1] [ 0 0 1 0] [ 0 1 0 0] 
They give representatives for $K_0(p^{n})\backslash Sp_4(F_p)$, so their first set of representatives reduces to just the identity, which is the same in either basis. Define the matrices of the following form where $z \in F_p$:
[1 0 0 0] [0 1 0 0] [0 z 1 0] [0 0 0 1] [1 0 0 0] [0 1 0 0] [0 z 1 0] [0 0 0 1] 
The second set of coset representatives are given by these matrices multiplied by $s_2$ on the right:
[ 1 0 0 0] [ 0 0 1 0] [ 0 1 z 0] [ 0 0 0 1] [ 1 0 0 0] [ 0 0 1 0] [ 0 1 z 0] [ 0 0 0 1] 
We now change this into our basis:
[1 0 0 0] [0 1 0 0] [0 0 1 0] [0 z 0 1] [1 0 0 0] [0 1 0 0] [0 0 1 0] [0 z 0 1] 
[ 1 0 0 0] [ 0 0 0 1] [ 0 0 1 0] [ 0 1 0 z] [ 1 0 0 0] [ 0 0 0 1] [ 0 0 1 0] [ 0 1 0 z] 
Now consider matrices of the following form for $y,z$ in $F_p$:
[1 0 0 0] [0 1 0 0] [y z 1 0] [0 y 0 1] [1 0 0 0] [0 1 0 0] [y z 1 0] [0 y 0 1] 
The representatives here are given by these matrices multiplied on the right by $s_2 s_1$. We change bases:
[1 0 0 0] [0 1 0 0] [0 y 1 0] [y z 0 1] [1 0 0 0] [0 1 0 0] [0 y 1 0] [y z 0 1] 
[ 0 1 0 0] [ 0 0 1 0] [ 0 0 y 1] [1 y z 0] [ 0 1 0 0] [ 0 0 1 0] [ 0 0 y 1] [1 y z 0] 
Finally, consider matrices of the following form for $x,y,z \in F_p$:
[1 0 0 0] [0 1 0 0] [y z 1 0] [x y 0 1] [1 0 0 0] [0 1 0 0] [y z 1 0] [x y 0 1] 
The representatives are given by these matrices multiplied on the right by $s_2 s_1 s_2$. We change bases:
[1 0 0 0] [0 1 0 0] [x y 1 0] [y z 0 1] [1 0 0 0] [0 1 0 0] [x y 1 0] [y z 0 1] 
[ 0 0 0 1] [ 0 0 1 0] [ 0 1 y x] [1 0 z y] [ 0 0 0 1] [ 0 0 1 0] [ 0 1 y x] [1 0 z y] 
