I don't think this works out for soft sheaves, and I think the following is an example: Let $X$ be the real line, and let $Z$ be the origin. Let $\mathcal F$ be the skyscraper sheaf on $X$ with stalk $\mathbb Z$ at the origin. I believe that should be soft (any section on any subset is determined by its germ at the origin and there is a unique global section for any such germ). Let $\mathcal G$ be the sheaf of germs of smooth real valued functions on $X$, which is known to be soft (see Bredon Example II.9.4). So, over $Z$, there is a map of the restricted sheaves determined by taking the generator of $\mathbb Z$ to a germ of a smooth function at the origin. But if that function germ is non-zero, its corresponding germs in nearby stalks will be non-trivial in $\mathcal G$ and so can't be in the images of the corresponding (trivial) germs of $\mathcal F$. In particular, the map can't be extended over even arbitrarily small neighborhoods of the origin and so can't be extended globally.

Sorry, this response is not quite as clearly written as I'd like, but does it make sense? If not, I'll try to clarify. (Another way to say this might be to consider the sheaves as equal to the presheaves of their sections. Then on any interval around the origin, the group of sections of $\mathcal F$ is $\mathbb Z$, so there are maps to the group of sections of $\mathcal G$ determined by where the generator goes. But if the image of the generator is represented by a nontrivial function on the interval (for example a non-zero constant function), this map won't commute with restrictions off the origin and so can't come from a sheaf map.)