$\delta$ is the transition function. It takes two arguments - the current state and a letter in the input alphabet - and its output is a subset of the states. This should contrast with your idea of a DFA, where the output is always a single state.
Try thinking about it this way: the transition function takes your current state and an input letter and returns a subset of $Q$, which could also be described as a binary number with $|Q|$ bits, each bit $i$ indicating whether state $i$ in or out of the subset.
Now, $\mathcal{P}(Q)$, the power set of $Q$, is defined to be the set of all subsets of $Q$. It also has a natural expression as the binary numbers with $|Q|$ bits or the functions from $Q$ to the set $\{0,1\}$. Note that the size of all of these is the number $2^{|Q|}$.
The last bit you're missing is that set theorists can construct the natural numbers by defining $2=\{0,1\}$. This may feel weird, but it's a useful shorthand here. Also, the set $A^B$ is defined to be the set of all functions from $B$ to $A$. Now, you can see the $2^Q$ in the definition of $\delta$ isn't a number at all, it's the set of functions from $Q$ to $2=\{0,1\}$, which is also naturally the power set of $Q$.
So, the domain of $\delta$ is $Q\times\Sigma$ and its range is $2^Q$ which is set-theoretically isomorphic to $\mathcal{P}(Q)$.
And you can't chain the transition function directly because it takes a state, not a set of states as input, but you can generalise it by defining $$\hat{\delta}(\hat{q},s) = \cup{\{\delta(q.s):q\in\hat{q}\}}$$ and then you can start at $\{q_0\}$ (rather than the state $q_0$) and apply $\hat{\delta}$ iteratively as you loop through your input word.
