We now prove that $|\mathcal{G}| \le 2^{m-1}(2^m + 1)^{k-1}$ for any $\mathcal{G}$ satisfying the conditions in the statement. For convenience, write $M = 2^m + 1$. Let $r_i$ denote the red element of $S_i$, and let $B_i$ be the set of blue elements in $S_i$.
For every subset $X_i \subset B_i$ and every $j \in \mathbb{Z}_M$, define the sets

$$T_{X_i, j} = \begin{cases} \{r_i\}, & \text{if } j = 0; \\ X_i, & \text{if } j \ne 0 \text{ and } j \text{ is even (considered as a number in } [1, M-1]\}; \\ B_i \setminus X_i, & \text{if } j \ne 0 \text{ and } j \text{ is odd (considered as a number in } [1, M-1]\}. \end{cases}$$

Note that, for every $i$ and every $j$, the sets $T_{X_i,j}$ and $T_{X_i,j+1}$ are disjoint. Now, for every sets $X_i \subset B_i$ and every elements $j_i \in \mathbb{Z}_M$, $i = 1, 2, \dots, k$, denote
$$F(X_1, X_2, \dots, X_k, j_1, j_2, \dots, j_k) = \bigcup_{i=1}^{k} T_{X_i, j_i}. \quad (*)$$

Claim. Every set $F \in \mathcal{F}$ has exactly $2^{mk}$ representations of the form ($*$).

Proof. Set $F_i = F \cap S_i$. If $F_i = \{r_i\}$, then there are $2^m$ possible choices for $X_i$, and one should necessarily have $j_i = 0$. Otherwise, there are only two possible choices for $X_i$, namely $X_i = F_i$ and $X_i = B_i \setminus F_i$, and for each of them there are $2^{m-1}$ possible choices for $j_i$. So, whatever $F$, there are $2^m$ possible choices for each pair $(X_i, j_i)$ all of which can be made independently, whence a total of $2^{mk}$ possible tuples $(X_1, X_2, \dots, X_k, j_1, j_2, \dots, j_k)$. This proves the Claim.

The Claim implies that each $F \in \mathcal{F}$ has the same number of representations of the form ($*$). Thus, it suffices to show that, among all $N = 2^{km}(2^m + 1)^k$ tuples $(X_1, X_2, \dots, X_k, j_1, j_2, \dots, j_k)$, at most $\frac{2^{m-1}}{2^m+1}N$ satisfy
$$F(X_1, X_2, \dots, X_k, j_1, j_2, \dots, j_k) \in \mathcal{G}.$$
To this end, split all these tuples into length $M$ cycles
$$(F(X_1, X_2, \dots, X_k, j_1, j_2, \dots, j_k), F(X_1, X_2, \dots, X_k, j_1+1, j_2+1, \dots, j_k+1), \dots, F(X_1, X_2, \dots, X_k, j_1+M-1, j_2+M-1, \dots, j_k+M-1)),$$
and note that any two adjacent sets of a cycle are disjoint. Hence each cycle contains at most $\lfloor M/2 \rfloor = 2^{m-1}$ sets from $\mathcal{G}$. This provides the desired upper bound and completes the solution.