Update: It seems I proved only the $(\dim S)$-decomposibility of the sum of two decomposables, since $s_1 + s_2$ is not decomposable (which I'm not quite sure of).
Original post:
Let $V$ be any $n$-dimensional real vector space and $\Lambda^k V$ be its $k$-th exterior power. Then I have this little conjecture about decomposability of $v \in \Lambda^k V$,
If $k > n/2 = \dim V/2$ then every (nonzero) $v \in \Lambda^k V$ is decomposable, i.e. can be written as a single product $v = v_1\wedge \dots \wedge v_k$ for some linearly independent vectors $v_1,\dots, v_k \in V$.
The motivation for that is that in $\mathbb R^3$ any 2-dimensional subspaces have a nontrivial intersection. That allows to rewrite each term in a sum $u_1\wedge u_2 + w_1\wedge w_2 = x\wedge (u_2 + w_2)$ for some $x \in S = U \cap W$, where $U = span\{u_1,u_2\}$, $W=span\{w_1,w_2\}$
In higher dimesnional spaces any subspaces whose dimension $k > n/2$ will necessarily intersect in a nontrivial subspace of dimension $\ge 2k - n > 0$. Recall that a decomposable $k$-vector corresponds to $k$-dimensional subspace. So any sum of two (and more) $k$-vectors $u,w$ will have there corresponding subspaces $U,W$ intersecting if $k > n/2$. In that subspace one can find a common $x \in \Lambda^{\dim S} V$ so that $ u + w = x \wedge s$, where $s \in \Lambda^{k-\dim S} V$.
Added:
The proof could be sketched as follows: each of the two decomposable $k$-vectors $u,w$ can be written as wedge of $k$ linearly independent vectors. Then intersection $S$ is a nontrivial subspace whose basis can be chosen as commen to both $u$ and $w$. Then $u = x\wedge s_1$ and $w = x \wedge s_2$ and $u + w = x \wedge (s_1 + s_2)$, where $x \in \Lambda^{\dim S} V$ and $s_1,s_2 \in \Lambda^{k - \dim S} V$.
What is known: if $n = \dim V$, then every $v \in \bigwedge^{n-1} V$ is decomposable.
Let's test your result on the case $n= 5$. Your claim is that every $v$ in $\bigwedge ^{3} V$ is decomposable.
Now, let's try to recall the Plucker relations in this case. They will be quadratic relations satisfied by $3\times 3$ minors of a $3\times 5$ matrix. First, for any $4$ columns of such a matrix we have (Cramer)
$$(1 2 3) 4 = (4 2 3)1+(143)2+(124)3$$
Now, add two more columns to get relations between minors. For instance
$$(123)(4 15)= (423)(115)+(143)(215)+(124)(315)$$ or $$(123)(145)+(134)(125)+(124)(135)=0$$ which is coming from an "old relation" for $\wedge^2(k^4)$. Nevertheless, it is a non-trivial relation.
It seems your conjecture is not true here.