Since $V^*$ is the initial vector space equipped with a homomorphism $V^* \otimes V \to K$, the question is closely connected to this question about geometric pictures for tensor products. But it also has no full answer yet.
I would suggest the following: If $K=\mathbb{R}$ and $V$ is a finite-dimensional vector space equipped with an inner product $\langle-,- \rangle$ (for example $V=\mathbb{R}^n$ with the standard product), then we may identify $V^*$ canonically with $V$ via $v \mapsto \langle v,-\rangle$. The operation $W \mapsto W^o$ corresponds to the orthogonal complement $W \mapsto W^{\perp}$, for which we have a good geometric picture. We have $W \subseteq W'$ iff $W'^\perp \subseteq W^\perp$, and $\dim(W^\perp)=\dim(V)-\dim(W)$. Thus, $\langle -,- \rangle$ makes it possible to define an "inner duality" of $V$ in the sense that $V$ becomes dual to itsself. For example, lines correspond to hyperplanes.
If $K$ is arbitrary and $V$ is a finite-dimensional vector space over $K$, then we have no canonical duality between $V$ and itsself. But $V$ is dual to another vector space, namely its dual vector space $V^*$ (this also motivates the terminology). Thus, there is a non-degenerate bilinear form $V^* \times V \to K$, namely $(\omega,v) \mapsto \omega(v)$ when you construct $V^*$ as $\hom_K(V,K)$. For $W \subseteq V$ we can still define the orthogonal complement $W^\perp \subseteq V^*$ with respect to this form. These orthogonal complements behave as above. Thus we may borrow our geometric intuition from the inner product case.
