Consider the field construction as in this link, where GF($2^m$) is obtained as a concatenation of additive subgroups of cardinality $2^j$ for $j=0,1,...,m$. In this question, no assumption is made that subgroups of cardinality $2^{j'}$ for $j'$ that divides $s$ need to be subfields.
Assume that two subgroups $\mathcal{S}_a$ and $\mathcal{S}_{b}$ are chosen (following the notations in the link above, where $\mathcal{S}_a$ is of cardinality $2^a$ and $\mathcal{S}_b$ is of cardinality $2^b$), where w.l.o.g. $\mathcal{S}_a \subseteq \mathcal{S}_b$ and $a+b \le m$. Can we always find an element $g \ne 0$ of the field such that $\{g \cdot {S_a}\}\bigcap {{S_b}} = \left\{ 0 \right\}$?
Note that $g \cdot {S_a}$ is an additive subgroup of GF($q$). Considering the subgroups as subspaces over GF($2$), my first thought was to look for $g$, such that when multiplied by the basis elements of $\mathcal{S}_a$, lead to a disjoint set of basis elements. However, I am not sure how to do this in a systematic way.
Addendum: For fixed values of $a,b$ such that $a+b \le m$ such $g$ can indeed be found. However, is there a "universal" value of $g$ such that $$\{ g \cdot {S_a}\} \bigcap {{S_b}} = \left\{ 0 \right\}$$ $$\{ {S_a}\} \bigcap {\left\{ {g \cdot {S_b}} \right\}} = \left\{ 0 \right\}$$ for every pair $\mathcal{S}_a,\mathcal{S}_b$ such that $a+b \le m$?
It is enough to consider $1\leq a$ and $a+b=m$, otherwise it becomes trivial case. Let $\{v_1, \ldots , v_a\}$ be a basis for $S_a$, and $\{w_1, \ldots, w_b\}$ be a basis for $S_b$. Let $F^*$ be the set of nonzero elements in $GF(2^m)$.
We find an element $g\in F^*$ satisfying: $$ g \cdot v_1 \notin S_b.\ \ \ (1) $$ Since multiplication by nonzero element is injective, there are at least $2^m - 1-2^b$ elements $g$ satisfy the above.
We also want $g$ to satisfy: $$ g\cdot v_2 \notin \mathrm{Span}(S_b \cup \{g\cdot v_1\}). \ \ \ (2) $$
The negation of this condition is in fact the existence of $a_{b+1}\in \{0,1\}$ such that $$ g\cdot v_2 = a_1 w_1 + \cdots + a_b w_b + a_{b+1} g\cdot v_1 $$
which is equivalent to $$ g \cdot (v_2 - a_{b+1} v_1) \in S_b. $$ We avoid all $g$ possibly satisfying the above. There are two choices for $a_{b+1}$ and $v_2 - a_{b+1} v_1 \neq 0$. Thus, multiplication by $v_2 - a_{b+1} v_1$ is injective. Thus, there are at least $2^m-1-2^b-2^{b+1}$ elements $g$ satisfy both (1) and (2).
We use this idea to find $g\in F^*$ satisfying (1) and:
For all $1\leq j\leq a-1$, $$ g\cdot v_{j+1} \notin \mathrm{Span}(S_b \cup \{g\cdot v_1 , \ldots , g\cdot v_j\}). \ \ \ (j+1) $$ Then the negation of the above is satisfied by at most $2^{j+b}$ elements $g\in F^*$. Therefore there are at least $2^m-1-\sum_{j=b}^{m-1} 2^j$ elements $g\in F^*$ satisfying (1) and ($j+1$) for all $1\leq j\leq a-1$.
We have $2^m-1-\sum_{j=b}^{m-1} 2^j\geq 1$. Thus, finding such $g$ is possible.