Can someone explain how the x and y jumped into the power, and somehow got the bitwise product operation performed on them?
It is helpful to write out a few examples to see what is happening.
1-bit Example:
$$
H|0\rangle
= \frac{1}{\sqrt{2}}|0\rangle + |1\rangle
= \frac{1}{\sqrt{2}}\sum_z |z\rangle (-1)^{0\cdot z}
$$
$$
H|1\rangle
= \frac{1}{\sqrt{2}}|0\rangle - |1\rangle
= \frac{1}{\sqrt{2}}\sum_z |z\rangle (-1)^{1\cdot z}
$$
2-bit Example:
$$
H|00\rangle
= \frac{1}{\sqrt{2^2}}(|0\rangle + |1\rangle)(|0\rangle + |1\rangle)
= \frac{1}{\sqrt{2^2}}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)
$$
$$
= \frac{1}{\sqrt{2^2}}\sum_z |z\rangle (-1)^{[00]\cdot z}
$$
$$
H|01\rangle
= \frac{1}{\sqrt{2^2}}(|0\rangle + |1\rangle)(|0\rangle - |1\rangle)
= \frac{1}{\sqrt{2^2}}(|00\rangle - |01\rangle + |10\rangle - |11\rangle)
$$
$$
= \frac{1}{\sqrt{2^2}}\sum_z |z\rangle (-1)^{[01]\cdot z}
$$
$$
H|10\rangle
= \frac{1}{\sqrt{2^2}}(|0\rangle - |1\rangle)(|0\rangle + |1\rangle)
= \frac{1}{\sqrt{2^2}}(|00\rangle + |01\rangle - |10\rangle - |11\rangle)
$$
$$
= \frac{1}{\sqrt{2^2}}\sum_z |z\rangle (-1)^{[10]\cdot z}
$$
$$
H|11\rangle
= \frac{1}{\sqrt{2^2}}(|0\rangle - |1\rangle)(|0\rangle - |1\rangle)
= \frac{1}{\sqrt{2^2}}(|00\rangle - |01\rangle - |10\rangle + |11\rangle)
$$
$$
= \frac{1}{\sqrt{2^2}}\sum_z |z\rangle (-1)^{[11]\cdot z}
$$
You see that, whatever the starting ket in our six above examples, and whether the number of qubits $n$ is 1 or 2, the expression on the far RHS is always the same, namely:
$$
H|x\rangle = \frac{1}{\sqrt{2^n}}\sum_z |z\rangle(-1)^{x\cdot z}\;.
$$
You can convince yourself that this is quite reasonable since wherever we have a single $|0\rangle$ ket in the direct product of the starting ket we get a
$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ and wherever we have a single $|1\rangle$ ket in the direct product of the starting ket we get a
$\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$. So, we can only pick up (and must pick up) minus signs in the expansion of the direct product when we expand through a $|1\rangle$.
If you want to prove it, try induction.