Polynomial kernelization for Set Splitting
I understood the question as &quot;select a subset of size $2k$ from $F$ such that (there exists coloring for $F$ $\iff$ there exists coloring for the subset)&quot;.
As you've said, you can remove all singleton and empty sets. The remaining sets have at least $2$ elements. You can show that any $2k$ such sets suffice.
Select $2k$ arbitrary sets with at least two elements. We assume that they have exactly $2$ elements (it's the worst case; the analysis is easy to modify for more than $2$ elements). We use the probabilistic method. For each element, we select the color randomly, and we compute the expected number of sets that have both colors.
We show that the expected number of sets with both colors is $k$. Note that this is an expectation over all random colorings. Therefore, there must exist a coloring that gives the value greater or equal to the expectation, i.e. $\ge k$, as required.
Let $\{S_i\}_{i=1}^{2k}$ be the selected sets.
For each set $S_i$ we introduce indicator variable $I_i$, which equals $1$ when the set $S_i$ has both colors, and $0$ otherwise.
The expectation above equals $\mathbb E[\sum_{i=1}^{2k} I_i] = \sum_{i=1}^{2k} \mathbb E[I_i]$.
For set $S_i = \{x_1, x_2\}$, with probability $\frac 12$, $x_1$ and $x_2$ have different colors. Therefore, $\mathbb E[I_i] = \frac 12$, and the required expectation is $2k \cdot \frac 12 = k$.