How many inputs does the Hadamard gate have?
The $n$-ary Hadamard gate acts on $n$ qubits. The state of the $n$ qubits is a unit norm vector of dimension $2^n$. You can take as basis the vectors $|x\rangle$ for $x \in \{0,1\}^n$.

As it happens, the $n$-ary Hadamard gate can be simulated by $n$ unary Hadamard gates acting on the individual qubits, which is what you see on page 6-3. The reason is that the $n$-dimensional Hadamard transform is the tensor product of $n$ one-dimensional Hadamard transforms.

The usual convention in circuit complexity is that gates have bounded fan-in. Otherwise you can compute any function using a large gate!

Consider the case $n=1$:
$$
\begin{bmatrix}
\frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}}
\end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix} =
\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} \, .
$$
This is where $|1\rangle$ comes from.

The entire circuit involves initializing $n$ qubits at the state $|0\rangle$, applying $n$ unary Hadamard gates on the $n$ qubits, applying the black box function $f$, and then applying the Hadamard gates again on the $n$ qubits. The final state of the $i$th qubit turns out to be $|s_i\rangle$.