Variant of (WEAK) PARTITION with 2 distinct solutions
No such reduction is likely to exist (unless P=NP).  In particular, there's a pseudo-polynomial time algorithm for your problem, so your problem isn't strongly NP-hard (unless P=NP).

The algorithm uses dynamic programming and is a straightforward modification of the dynamic programming algorithm for PARTITION.  Define $f(j,y,z)$ to be 1 if there exists two non-zero disjoint vectors $x,x'$ of the required form such that

$$\sum_{i=1}^j x_i a_i = y, \sum_{i=1}^j x'_i a_i = z.$$

Otherwise, $f(j,y,z)=0$.  Also define $g(j,y)$ to be 1 if there exists a non-zero vector $x$ such that $\sum_{i=1}^j x_i a_i = y$, or 0 otherwise.

Note that $f,g$ satisfy the following recurrence relations:

$$\begin{align*}
f(j,y,z) = &amp;f(j-1,y,z) \lor f(j-1,y-a_j,z) \lor f(j-1,y+a_j,z) \lor f(j-1,y,z-a_j)\\
&amp;\lor f(j-1,y,z+a_j) \lor (g(j-1,y) \land z=\pm a_j) \lor (g(j-1,z) \land y=\pm a_j).\\
g(j,y) = &amp;g(j-1,y-a_j) \lor g(j-1,y+a_j) \lor g(j-1,y) \lor (y=\pm a_j).\end{align*}$$

The base cases are $f(0,y,z)=0$ and $g(0,y)=0$ for all $y,z$.

With this recurrence, evaluate $f(j,y,z)$ and $g(j,y)$ for all $0 \le j \le n$ and all $y,z$ satisfying $|y|,|z| \le a_1 + \dots + a_n$.  Each such value can be computed in $O(1)$ time using memoization / dynamic programming.

Finally, look at the value of $f(n,0,0)$ that you computed.  If $f(n,0,0)=1$, then there is a solution to your problem (i.e., there exists two disjoint vectors $x,x'$ that each represent two disjoint weak partitions).

What's the running time of the algorithm? If the inputs are represented in unary, the number of $f$-values we compute is polynomial in the size of the input, and each $f$-value can be computed in $O(1)$ time using the recurrence relation above.  It follows that this dynamic programming algorithm runs in polynomial time, if the inputs are provided  in unary.