Given an $n$-dimension vector space $V$ over a finite field $\mathbb F_q$ and a natural number $d<n$, the goal is to write $V$ as disjoint union of $d$-dimensional affine subspaces $v_i+V_i$: $$V = \stackrel{.}{\cup}_i (v_i + V_i)\quad \mbox{ where } v_i\in V \mbox{ and } V_i\le V \mbox{ with } \dim_{\mathbb F_q} V_i = d$$
One way to obtain such a partition with not all $V_i$ being the same is to pick a hyperplane $H\le V$ (i.e., $\dim_{\mathbb F_q} H = n-1$), write $V = \stackrel{.}{\cup}_i (v_i + H)$, and refine this partition by partitioning each $v_i+H$ independently in a recursive way.
Are there partitions that cannot be derived in this way?
(If necessary you may assume $q=2$, as this is the case I'm mostly interested in.)
No. With $q=2$ every two element subset $\{x,y\}$ is a coset of the subspace $\{0,x-y\}$. And you can partition a 3-dimensional space over $\Bbb{F}_2$ into four such doublets in such a way that no two of them form an affine 2-space. For example $$ \{000,100\}, \{010,011\}, \{001,110\}, \{111,101\}. $$