The question is about the decomposition of a 10d spinor into 4d and 6d pieces. The relevant fact is that under the group reduction $Spin(1,9) \rightarrow Spin(1,3) \times Spin(6)$, the 10d Weyl spinor of positive chirality $\mathbf{16}_+$ decomposes as $(\mathbf{2}_+ , \mathbf{4}_+) \oplus (\mathbf{2}_- , \mathbf{4}_-)$. Here, $\mathbf{n}$ denotes a representation of complex dimension $n$ and the indices $\pm$ refer to the positive/negative chirality.

The representations $\mathbf{2}_+$ and $\mathbf{2}_-$ are the two usual Weyl spinors in $4d$. The representations $\mathbf{4}_+$ and $\mathbf{4}_-$ are the two Weyl spinors representations of $Spin(6)$ (they are the fundamental and antifundamental of $SU(4) = Spin(6))$. In general, it is useful to know that a Dirac spinor in dimension $N$ has $2^{ [N/2] }$ complex components, where $[ ]$ denotes the integer part, and that when $N$ is even, there exists Weyl spinors having $\frac{1}{2} 2^{ [N/2] }$ complex components.

To guess the form of the decomposition of $\mathbf{16}_+$, the first thing to do is to remark that the 10d chirality operator is the product of the 4d and 6d chirality operators (it is an easy exercise using the expressions of chirality operators in terms of gamma matrices). This implies that in the decomposition, the spinors with respect to $Spin(1,3)$ and $Spin(6)$ have to have the same chirality and so there are two possibilities : $(+,+)$ or $(-,-)$. This is probably the origin of the "ansatz". Then dimension considerations, using the countings of the preceding paragraph, suggest the answer.

This reasoning is more a way to quickly guess the answer. If one wants to rigorously prove the result, one has to know how to construct the various spin representations. This is done for example in the Appendix B of volume 2 of Polchinski's book on string theory.