$f\in\mathrm{Hom}(\nu,\mu)$ means that $f$ is a homomorphism with domain $\nu$ and codomain $\mu$.
It is clear later that the arguments of $f$ are $p$-tuples of the form $(a_1,\ldots,a_p)$. Being "homogeneous" means that there exists a natural number $k$ such that for all scalars $\alpha$, $f(\alpha a_1,\alpha a_2,\ldots,\alpha a_p) = \alpha^k f(a_1,a_2,\ldots,a_p)$. The natural number $k$ is the degree of this homogeneous map.
Just like the "inputs" to $f$ are tuples, the outputs are meant to be $n$-tuples. Each component is computed separately, so you can think of each component as consisting of a function $f_i(a_1,\ldots,a_p)$. For example, if you have a function $f\colon\mathbb{R}^2\to\mathbb{R}^3$ given by $f(x,y) = (x^2-y, x+y, y-2x)$, then you can think of $f$ as consisting of three "component" functions, $f_1$, $f_2$, and $f_3$, with $f_1(x,y)=x^2-y$, $f_2(x,y)=x+y$, and $f_3(x,y)=y-2x$. Then we write $f=(f_1,f_2,f_3)$. That's what the notation $f=(f_1,\ldots,f_m)$ means.
$A^p$ is the collection of all $p$-tuples of elements of $A$.
Frankly, I don't think you are meant to make sense of that quote. For one thing, the "quote" misuses "$o$" to stand for $\circ$, composition, writing $f_iog$ instead of $f_i\circ g$. Which suggests the author of the book didn't really know what he was quoting either.