Does fast function forcing really have $\kappa$-Knaster property?

I agree with you; I think your antichain example shows that fast function forcing is not $\kappa$-c.c.

Since I am a little surprised to hear that there has been a mistake about this in the literature, I wonder whether they were talking about some modified version of the fast function forcing? For example, I believe that if one modifies the forcing to insist that $p(\gamma)<\gamma$, then I believe that one could run a $\Delta$-system argument and show that it has the $\kappa$-c.c. In this case, however, the generic function would not be so fast, and so the "fast function" moniker would no longer be apt.

Meanwhile, let me also mention that there are several different versions of fast function forcing mentioned in the literature. The definition you give is one that I also used in an early paper, but I came eventually to realize that it is not as good with factoring as the definition that requires the smallness property $|p\upharpoonright\lambda|<\lambda$ at every inaccessible cardinal $\lambda$, rather than just at $\lambda\in\text{dom}(p)$. I now always use this latter version of fast function forcing. The two forcing notions are different, because one could have a condition $p$ with $p\upharpoonright\lambda$ having domain unbounded in an inaccessible cardinal $\lambda$, provided only that $\lambda\notin\text{dom}(p)$. This can cause some irritating issues with factoring the forcing, and it is more natural to avoid it.

But for both these kinds of fast-function forcing, the same simple antichain that you mention shows it is not $\kappa$-c.c. Indeed, to my way of thinking, the main point of fast function forcing is to arrange that $j(f)(\kappa)$ can be anything you want up to $j(\kappa)$, and the conditions forcing these various values of course form a large antichain; so the lack of the $\kappa$-c.c. seems to be inherent in fast function forcing.

As for question 2, the answer is clearly yes, if you allow $h$ in $V[f]$, since the range of $f$ is unbounded in $\kappa$, and so you can just let $h(\alpha)$ be an ordinal $\beta$ for which $g(\alpha)<f(\beta)$. But do you want $h\in V$? In this case, the answer is negative, since one can take $g=f$ itself, and there there will be no $h\in V$ with $f(\alpha)<f(h(\alpha))$, since if a condition $p$ forced that $h$ worked, we could extend $p$ be specifying it at $\alpha$ and $h(\alpha)$ in such a way that violated the requirement.