$V = \bigoplus_{i<n} U_i$ if and only if every $v \in V$ is $\sum_{i<n} u_i$ with $u_i \in U_i$ in exactly one way; equivalently, if and only if the sum is $V$ and $\sum_{i<n} u_i = 0_V$ with $u_i \in U_i$ forces every $u_i = 0_V$

Let $u$ be admissible and $j < n$. The list $u^{(j)}$ is admissible for the family $U^{(j)}$, since $u^{(j)}_i = u_i \in U_i = U^{(j)}_i$ for $i \ne j$ and $u^{(j)}_j = 0_V \in \{0_V\} = U^{(j)}_j$; hence $s := \sum_{i<n} u^{(j)}_i$ lies in $\sum_{i \ne j} U_i$, and $\sum_{i<n} u_i = u_j + s$.

A linear subspace $W$ of $V$ is closed under additive inverses: for $x \in W$ we have $(-1_F)x \in W$ by closure under scalar multiplication, and $(-1_F)x = -x$.

Let $j < n$ and $x \in U_j$. The list $z : n \to V$ with $z_j = x$ and $z_i = 0_V$ for $i \ne j$ is admissible, each $U_i$ containing $0_V$, and it sums to $x$.

(c) implies (b). Assume (c). Existence: since $\sum_{i<n} U_i = V$, every $v \in V$ is $\sum_{i<n} u_i$ for some admissible $u$. Uniqueness: suppose $u$ and $w$ are admissible with $\sum_{i<n} u_i = \sum_{i<n} w_i =: v$. The mixed identity with $\lambda = -1_F$, applied to $w$ and $u$ in that order, gives $(-1_F)\sum_{i<n} w_i + \sum_{i<n} u_i = \sum_{i<n}\bigl((-1_F)w_i + u_i\bigr)$, whose left-hand side is $-v + v = 0_V$; the list $i \mapsto (-1_F)w_i + u_i = (-w_i) + u_i$ is admissible, each $U_i$ being closed under additive inverses and addition; so by (c) every $(-w_i) + u_i = 0_V = (-w_i) + w_i$, and cancelling $-w_i$ on the left gives $u_i = w_i$ for every $i < n$.

(a) implies (c). Assume (a). Condition (D1) is the first half of (c). For the second, let $u$ be admissible with $\sum_{i<n} u_i = 0_V$ and let $j < n$. Writing $s = \sum_{i<n} u^{(j)}_i \in \sum_{i \ne j} U_i$, we get $u_j + s = 0_V$, while $(-s) + s = 0_V$ as well, so cancelling $s$ on the right gives $u_j = -s$; and $-s \in \sum_{i \ne j} U_i$ because that set is a linear subspace. Hence $u_j \in U_j \cap \sum_{i \ne j} U_i$, which is $\{0_V\}$ by (D2), so $u_j = 0_V$. As $j < n$ was arbitrary, $u$ is the all-zero list.

(b) implies (a). Assume (b). For (D1): every $v \in V$ is $\sum_{i<n} u_i$ for some admissible $u$, so $V \subseteq \sum_{i<n} U_i$, and the reverse inclusion holds because $\sum_{i<n} U_i$ is a subset of $V$. For (D2): let $j < n$ and $x \in U_j \cap \sum_{i \ne j} U_i$. Then $x = \sum_{i<n} w_i$ for some list $w$ admissible for $U^{(j)}$; such a $w$ has $w_j = 0_V$ and $w_i \in U_i$ for $i \ne j$, so it is admissible for $U$ as well, $U_j$ containing $0_V$. The list $z$ of step 1.3 is also admissible and also sums to $x$, so uniqueness in (b) forces $z = w$, and in particular $x = z_j = w_j = 0_V$. Since $\{0_V\}$ is contained in the intersection anyway, (D2) holds.

Steps 2.1, 1.4 and 2.2 give (a) implies (c), (c) implies (b) and (b) implies (a), so the three conditions are equivalent.