The result is true indeed. Here is a rather technical solution. I work by induction over $k$, thus denoting by $S_k$ the above $(2k)\times(2k)$ matrix. Let us transform the matrix through the following series of row and column operations:
$C_{2k} \leftarrow C_{2k}+s_1 C_{2k-1}$ and
then, for *odd* $j$ from $3$ to $2k-3$,

$L_k \leftarrow L_k+s_{k-j}L_j$ for $k$ from $j+1$ to $2k$.

After those operations, the matrix obtained from $S_k$ has the following form:
$$S'_k=\begin{bmatrix}
0 & D \\
T & ? \end{bmatrix}
$$
where $D=\begin{bmatrix}
1 & 0 \\
s_1 & 0
\end{bmatrix}$, the odd-labelled columns of
$T$ are $\begin{bmatrix}
0 \\
0 \\
\vdots \\
0 \\
0 \\
1 \\
0
\end{bmatrix}$, $\begin{bmatrix}
0 \\
0 \\
\vdots \\
1 \\
0 \\
0 \\
0
\end{bmatrix}$, ..., $\begin{bmatrix}
1 \\
0 \\
\vdots \\
0 \\
0 \\
0 \\
0
\end{bmatrix}$, and $T$ is equivalent to $S_{k-1}$.

By the induction hypothesis $T$ has rank $k-1$ and hence its column space is spanned by the above vectors. To conclude that $S_k$ has rank $k$ we then need to prove that the even-labelled entries in the last column of $S'_k$ are all zero.

Let $p \in \{2,...,k\}$. By careful examination, one finds that the $(2p,2k)$-entry of $S'_k$ equals
$$s_{2p}+\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}$$
where $A_p$ is the set of all lists $(i_1,\dots,i_\ell)$ of positive integers whose sum is
$2p$ and in which $i_1$ and $i_\ell$ are odd and the other entries are even.
Now, we show by induction that this sum equals $s_{2p}$. First of all, $\theta : (i_1,\dots,i_\ell) \mapsto (i_\ell,\dots,i_1)$ is an involution on $A_p$, and as $\mathbb{F}$ has characteristic $2$ we deduce that
$$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}=
\sum_{(i_1,\dots,i_\ell) \in A_p,(i_1,\dots,i_\ell)=(i_\ell,\dots,i_1)} s_{i_1} s_{i_2}\cdots s_{i_l}.$$

Next, if we have a symmetric list $(i_1,\dots,i_{2j})=(i_{2j},\dots,i_1)$ in $A_p$ with an even number of entries and $j \geq 2$, we use equality $s_{i_j}s_{i_{j+1}}=s_{i_j}^2=s_{2i_j}$ to see that it defines the same product has the symmetric list $(i_1,\dots,i_{j-1},2i_j,i_{j+2},\dots,i_{2j})$ with $2j-1$ entries.
Pairing lists in this manner and - if $p$ is odd - noting that $s_p^2=s_{2p}$, we find that
$$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}=
\begin{cases}
s_{2p}+\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is odd} \\
\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is even,}
\end{cases}
$$
where $B_p$ denotes the subset of $A_p$ consisting of the symmetric lists
$(i_1,\dots,i_{2j+1})$ in which $i_{j+1}=2t$ for some odd integer $t$.
If $p$ is odd then $B_p$ is empty and we are done.

Assume now that $p=2q$ for some integer $q$. Then,
$(i_1,...,i_{2j+1}) \mapsto (i_1,...,i_{j-1},i_j,(i_{j+1})/2)$
maps $B_p$ bijectively onto $A_q$,
and using $s_{i_{j+1}}=s_{(i_{j+1}/2)}^2$ we deduce that
$$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l}
=\Bigl(\sum_{(i_1,\dots,i_\ell) \in A_q} s_{i_1} s_{i_2}\cdots s_{i_l}\Bigr)^2.$$
By induction, we deduce that
$$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l}
=(s_{2q})^2=s_{2p},$$
QED.