Let $X$ be a simplicial set and let $\wedge_i^2$ be the $i-th$ horn of the simplicial set $\Delta^2$. The Kan condition is the horn filling condition for $i=0,1,2$ and the weak Kan condition is the horn filling condition for $i=1$. In the setting of the diagram 
we let $f$, $g$, $h$ be assignments of 1-simplices in $X$ satisfying the obvious compatibility conditions. The Kan condition for $i=0$ says that $g \circ f$ can be defined.
Question: Taking $h$ to be the null degenerate 'identity' assignment, and using the Kan condition for $i=2$, we get that there is a candidate for an inverse of $g$. Is there any sense in which this implies that there is a unique extension of any horn?(In other words, is the difference between weak and regular Kan complexes the uniqueness of the extension of a horn.)
I am already seeing some difficulties with my characterization in the case of the singular set of a topological space. But I just want to find out if the difference between weak and regular Kan complexes can be summarized in terms of something like this.
If I recall and understand your question correctly, horn fillers are not necessarily unique. However, for (weak) Kan complexes they are unique up to contractible choice.
One way to think about weak Kan complexes and Kan complexes is as follows:
Kan complexes are $\infty$-groupoids, i.e., $(\infty, 0)$-categories. All morphisms are invertible, but "composition" is not unique. One can also think of them as topological spaces, and there are multiple ways of composing paths.
A weak Kan complex with unique inner horn fillers is isomorphic to the nerve of some category. Not all morphisms are invertible, but there is a unique composition law.
In general, weak Kan complexes are $(\infty, 1)$-categories (tautologically, depending on your model). Not all $1$-morphisms are invertible (but all $n$-morphisms, $n > 1$, are) and composition is not unique.
In a sense, weak Kan complexes are a simultaneous generalization of topological spaces and ordinary categories.