The secret polynomials are multivariate polynomials whose oil-oil terms have coefficient zero. You can represent it as a sum of terms, or as a vector-matrix-vector product. Take for example the polynomial of your example, $F_0(x_0, \ldots, x_5) = x^2_0 − 4x_0x_1 − 4x^2_1 + 8x_0x_2 + 11x_2^2 − 2x_0x_3 − x_1x_3 + 9x_2x_3 + 12x^2_3 + 7x_0x_4 − 11x_1x_4 − 9x_2x_4 − 2x_3x_4+3x_0x_5+5x_1x_5 + 14x_2x_5 − 11x_3x_5$. Using the vector notation $\mathbf{x}^\mathsf{T} = (x_0, x_1, \ldots, x_5)$ we can write this polynomial as the following product:
$$ F_0(\mathbf{x}) = \mathbf{x}^\mathsf{T} \left( \begin{matrix}
1 & -4 & 8 & -2 & 7 & 3 \\
0 & -4 & 0 & -1 & -11 & 5 \\
0 & 0 & 0 & 9 & -9 & 14 \\
0 & 0 & 0 & 12 & -2 & -11 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0
\end{matrix} \right) \mathbf{x} \enspace .$$
The coefficients in the matrix are the same as in the sum-of-terms representation. Note that in this matrix representation, the bottom-right block consists of zeros; this reflects the fact that oil-times-oil terms have coefficient zero. Also note that this matrix is not unique: whenever $F_i(\mathbf{x}) = \mathbf{x}^\mathsf{T} M_{F_i} \mathbf{x}$ then for any skew-symmetric matrix $A$ (i.e. such that $A^\mathsf{T} = -A$) you can also write $F_i(\mathbf{x}) = \mathbf{x}^\mathsf{T}(M_{F_i}+A)\mathbf{x}$. In particular, this means that (assuming odd characteristic) it is always possible to choose the central matrix to be symmetric by computing $\frac{1}{2}(M_{F_i}+M_{F_i}^\mathsf{T})$.
The secret linear transform $\mathcal{T}$ can also be represented in two ways. You give the matrix representation
$$ T = \left( \begin{matrix}
18 & 23 & 14 & 1 & 6 & 21 \\
24 & 3 & 3 & 0 & 2 & 0 \\
17 & 25 & 2 & 16 & 23 & 8 \\
8 & 21 & 16& 5 & 1 & 9 \\
14 & 28 & 8 & 17 & 12 & 12 \\
6 & 24 & 18 & 19 & 3 & 1
\end{matrix} \right) \enspace ,$$
in which case $\mathcal{T}(\mathbf{x}) = T\mathbf{x}$. You could also have chosen the list of polynomials representation, in which case there are 6 polynomials, $\mathcal{T}_0$ throught $\mathcal{T}_5$. For illustration, the first polynomial is given by $\mathcal{T}_0(x_0, x_1, x_2, x_3, x_4, x_5) = 18 x_0 + 23 x_1 + 14 x_2 + 1 x_3 + 6 x_4 + 21 x_5$.
To compute the composition $F \circ \mathcal{T}$ you must first choose a representation. In the polynomial representation we have $F \circ \mathcal{T} (\mathbf{x}) = F(\mathcal{T}(\mathbf{x})) = F(\mathcal{T}_0(\mathbf{x}), \mathcal{T}_1(\mathbf{x}), \ldots, \mathcal{T}_5(\mathbf{x}))$. So you can compute $F \circ \mathcal{T}$ by starting with a copy of $F$ and then substituting every occurrence of $x_i$ with $\mathcal{T}(\mathbf{x}) = \mathcal{T}(x_0, x_1, x_2, x_3, x_4, x_5)$. But make sure to keep track of which occurrences of $x_i$ appear as a result of the substitution, because they obviously shouldn't be substituted twice.
In the matrix representation, we have $F_i \circ \mathcal{T} (\mathbf{x}) = \mathcal{T}(\mathbf{x})^\mathsf{T} M_{F_i} \mathcal{T}(\mathbf{x}) = (T\mathbf{x})^\mathsf{T} M_{F_i} (T \mathbf{x}) = \mathbf{x}^\mathsf{T} (T^\mathsf{T} M_{F_i} T) \mathbf{x}$. So a matrix representation $M_{F_i \circ \mathcal{T}}$ of the composition can be computed just by sandwiching $M_{F_i}$ between $T^\mathsf{T}$ and $T$, i.e., $M_{F_i \circ \mathcal{T}} = T^\mathsf{T} M_{F_i} T$.
To complete the example you started, let's assume we're working modulo 31 (because otherwise the numbers get quite large). Then we have
$$ M_{F_0 \circ \mathcal{T}} \cong T^\mathsf{T} M_{F_i} T = \left( \begin{matrix}
6 & 25 & 19 & 19 & 3 & 26 \\
22 & 29 & 3 & 29 & 28 & 2 \\
8 & 15 & 12 & 18 & 6 & 7 \\
28 & 1 & 24 & 17 & 4 & 1 \\
6 & 21 & 18 & 5 & 30 & 23 \\
11 & 23 & 8 & 15 & 5 & 5
\end{matrix}\right) \cong \left( \begin{matrix}
6 & 8 & 29 & 8 & 20 & 3 \\
8 & 29 & 9 & 15 & 9 & 28 \\
29 & 19 & 12 & 21 & 12 & 23 \\
8 & 15 & 21 & 17 & 20 & 8 \\
20 & 9 & 12 & 20 & 30 & 14 \\
3 & 28 & 23 & 8 & 14 & 5
\end{matrix}\right) \cong \left( \begin{matrix}
6 & 16 & 27 & 16 & 9 & 6 \\
0 & 29 & 18 & 30 & 18 & 25 \\
0 & 0 & 12 & 11 & 24 & 15 \\
0 & 0 & 0 & 17 & 9 & 16 \\
0 & 0 & 0 & 0 & 30 & 28 \\
0 & 0 & 0 & 0 & 0 & 5
\end{matrix}\right) \enspace .$$
Here the congruence sign ($\cong$) indicates that the matrices represent the same quadratic form, i.e., up to addition of skew-symmetric matrices. In other words, after sandwiching them between $\mathbf{x}^\mathsf{T}$ and $\mathbf{x}$ they will result in the same polynomial. From the last upper triangular matrix we can read out the representation of the polynomial as a sum of terms, namely
$$ F_0 \circ \mathcal{T}(x_0, x_1, x_2, x_3, x_4, x_5) = 6x_0^2 + 16x_0x_1 + 27x_0x_2 + 16x_0x_3 + 9x_0x_4 + 6x_0x_5 + 29x_1^2 + 18x_1x_2 + 30x_1x_3 + 18 x_1x_4 + 25x_1x_5 + 12x_2^2 + 11x_2x_3 + 24x_2x_4 + 15x_2x_5 + 17x_3^2 + 9x_3x_4 + 16x_3x_5 + 30x_4^2 + 28x_4x_5 + 5x_5^2 \enspace .$$
This should be the same result as obtained through the substitution method applied to the polynomial representations.
One more thing: it is a good idea to avoid mixing homogeneous quadratic polynomials (i.e., without linear or constant terms) with affine transforms (i.e., with constant terms) -- or vice versa. In other words, whenever your secret polynomials are homogeneous, your secret transforms should be linear; whenever they are inhomogeneous, they should be affine. This guarantees that the public polynomials are (in)homogeneous whenever the secret ones are.
