\noindent
We will show that there are no free morphisms of degrees $4$ and $5$ into $X$.
The morphism $\varphi = (G_0, ..., G_5)$, where $G_i = \sum_{0 \leq j \leq d}a_{ij}S^{d - j}T^{j}$ are homogeneous polynomials of degree $d$, gives us the $6 \times (d+1)$ matrix $(a_{ij})$. We will denote this matrix as $M_{\varphi}$. Note that the transpose, $M^{T}_{\varphi}$, is the matrix of the $k$-linear map $\widetilde{\varphi}_d: (R(-d)^{\oplus 6})_d \rightarrow R_d$. We will also need: (i) $f_1 + ... + f_5 = d$, (ii) $f_l = 4e_l + 5d$ for $l = 1, ..., 5$, and (iii) There are no degree $d$ free morphisms into $X$ which are not $E$-very free, where $4$ does not divide $d$.\\

\noindent
\begin{lemma} If $\varphi$ is a degree $4$ or $5$  free morphism into $X$, then $M_{\varphi}$ has maximal rank.
\end{lemma}

\noindent
\begin{proof} $Degree$ $4$: The $R$-linear map of graded free modules $\widetilde{\varphi}: R(-4)^{\oplus 6} \xrightarrow{(G_0,...,G_5)} R$, induces a $k$-linear map $\widetilde{\varphi}_4: (R(-4)^{\oplus 6})_4 \rightarrow R_4$ on the degree $4$ graded pieces.
Since dim$_k((R(-4)^{\oplus 6})_4) = 6$, and dim$_k(R_4) = 5$, $ker(\widetilde{\varphi}_4)$ is nontrivial. 

\noindent
If dim$_k(ker(\widetilde{\varphi}_4)) \geq 2$, there exists distinct $i, j \in \{1,2,3,4,5\}$ such that $e_i = -4 = e_j$. By (ii), $f_i = 4 = f_j$, and by assumption, $f_l \geq 0$ for all $l$. So (i) $\Rightarrow$ we must have some $f_l < 0$, a contradiction. Hence, dim$_k(\widetilde{\varphi}_4)) = 1$, and $M_{\varphi}$ has maximal rank.\\

\noindent
$Degree$ $5$: Consider the $k$-linear map $\widetilde{\varphi}_5: (R(-5)^6)_5 \rightarrow R_5$. We have dim$_k((R(-5)^6)_5)$ $= 6 =$ dim$_k(R_5)$. Suppose $M_{\varphi}$ does not have maximal rank. Then $ker(\widetilde{\varphi}_5)$ is nontrivial, and there exists some $e_i = -5$, say w.l.o.g. $e_1$. So, $f_1 = 5$, and by (i), $f_2 = f_3 = f_4 = f_5 = 0$. But this is impossible by (iii). This shows that $M_{\varphi}$ has maximal rank, and being a square matrix, it is invertible. 
\end{proof}

\noindent
\begin{lemma} (a) There are no degree $4$ free morphisms into $X$.

\hspace{18mm}(b) There are no degree $5$ free morphisms into $X$.
\end{lemma}

\noindent
\begin{proof} $(a)$ Assume a free morphism $\varphi = (G_0, ..., G_5)$ exists, where for $i = 0,..., 5, G_i = \sum_{0 \leq j \leq 4}a_{ij}S^{4 - j}T^j$. By the previous lemma, the matrix $M_{\varphi} = (a_{ij})$ has maximal rank. Since permuting the $G_i's$ does not affect the splitting type of $E_X(\varphi)$, we can assume that the first $5$ rows of $M_{\varphi}$ are linearly independent over $k$. So, $det((a_{ij})_{i \leq 4}) \neq 0$.\\

\noindent
Consider the matrix $\overline{M_{\varphi}} = (a^4_{ij})$. As char$(k) = 2$, the map $\phi: k \rightarrow k$ given by $\phi(a) = a^4$ is a ring homomorphism. But this means that $det((a^4_{ij})_{i \leq 4}) = (det((a_{ij})_{i \leq 4}))^4 \neq 0$, proving that $\overline{M_{\varphi}}$ has maximal rank.\\

\noindent
A degree $4$ homogeneous polynomial in $R$ is of the form $a_0S^4 + a_1S^3T + a_2S^2T^2 + a_3ST^3 + a_4T^4$. If we denote the coefficient of a term of a polynomial in $R$ as $cf(term)$, then raising the degree $4$ homogeneous polynomial to the $5^{th}$ power, we get the following:\

\noindent
$cf(S^{19}T) = a^4_0a_1$; $cf(S^{15}T^5) = a^4_1a_1$; $cf(S^{11}T^9) = a^4_2a_1$; $cf(S^7T^{13}) = a^4_3a_1$; $cf(S^3T^{17}) = a^4_4a_1$; $cf(S^{17}T^3) = a^4_0a_3$; $cf(S^{13}T^7) = a^4_1a_3$; $cf(S^9T^{11}) = a^4_2a_3$; $cf(S^5T^{15}) = a^4_3a_3$; $cf(ST^{19}) = a^4_4a_3$.\\

\noindent
Since $G^5_0 + ... + G^5_5 = 0$, for $j = 0,1,2,3,4$, we get $\sum_{0 \leq i \leq 5}a^4_{ij}a_{i1} = 0$, and similarly, $\sum_{0 \leq i \leq 5}a^4_{ij}a_{i3} = 0$. $(*)$\\

\noindent
Consider the linear map $k^6 \rightarrow k^5$ given by right multiplication by the matrix $\overline{M_{\varphi}}$. The kernel of this linear map has dimension $1$, because rank$(\overline{M}_{\varphi}) = 5$. By $(*)$, $(a_{01}, a_{11}, ...,a_{51}), (a_{03}, a_{13}, ..., a_{53}) \in ker(k^6 \rightarrow k^5)$, and these $6-$tuples are also columns of $M_{\varphi}$, hence linearly independent over $k$. This means that $ker(k^6 \rightarrow k^5)$ has dimension at least $2$, a contradiction. This proves $(a)$.\\

\noindent
$(b)$ Assume $\varphi = (G_0, ..., G_5)$ is a degree $5$ free morphism, where $G_i = \sum_{0 \leq j \leq 5}a_{ij}S^{5-j}T^j$. Then by the previous lemma, $M_{\varphi} = (a_{ij})$ is invertible. It follows that $\overline{M_{\varphi}} = (a^4_{ij})$ is also invertible.\\

\noindent
Raising a degree $5$ homogeneous polynomial $a_0S^5 + a_1S^4T + a_2S^3T^2 + a_3S^2T^3 + a_4ST^4 + a_5T^5 \in R$ to the $5^{th}$ power, we get the following:\

\noindent
$cf(S^{23}T^2) = a^4_0a_2$; $cf(S^{19}T^6) = a^4_1a_2$; $cf(S^{15}T^{10}) = a^4_2a_2$; $cf(S^{11}T^{14}) = a^4_3a_2$; $cf(S^7T^{18}) = a^4_4a_2$; $cf(S^3T^{22}) = a^4_5a_2$.\\

\noindent
Since $G^5_0 + ... + G^5_5 = 0$, for $j= 0, 1, 2, 3, 4, 5$, we get $\sum_{0 \leq i \leq 5}a^4_{ij}a_{i2} = 0$. $(**)$\\

\noindent
Now, the $k$-linear map $k^6 \rightarrow k^6$ given by right multiplication by $\overline{M_{\varphi}}$ is an isomorphism. By $(**)$, $(a_{02}, a_{12}, ... , a_{52}) \in ker(k^6 \rightarrow k^6)$, and the invertibility of $M_{\varphi}$ implies that $(a_{02}, ..., a_{52}) \neq (0, ..., 0)$. This contradicts the triviality of $ker(k^6 \rightarrow k^6)$.
\end{proof}
