First note that $H = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} $ and $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $ and $|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

Hence $H|0\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}} $

Thus, we have that:
$$\big( H\otimes I \big) |00 \rangle = H|0\rangle \otimes I|0\rangle = \bigg(\dfrac{|0\rangle + |1\rangle}{\sqrt{2} } \bigg)\otimes |0\rangle = \dfrac{|00\rangle + |10\rangle}{\sqrt{2} } $$

Now note that $CNOT = |0\rangle\langle 0| \otimes I +|1 \rangle \langle 1| \otimes X =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix} $. Thus, the CNOT gate here just apply the bit flip gate, $X$ gate, to the target qubit when the controlled-qubit is $|1\rangle$ and do nothing otherwise.

Therefore, if you apply $U = CNOT \cdot H\otimes I$ to the state $|S\rangle = |00\rangle$ then you get
$$ \big( CNOT \cdot H\otimes I\big) |00\rangle = CNOT \cdot \bigg( \big(H \otimes I \big) |00\rangle \bigg) = CNOT \cdot \bigg(\dfrac{|00\rangle + |10\rangle}{\sqrt{2} } \bigg) = \dfrac{|00\rangle + |11\rangle}{\sqrt{2} } $$

That is, $U |00\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2} } $. On a quantum circuit, this can be realized as:

Note that for the figure above, $|\psi_0 \rangle = |S\rangle = |00\rangle$, $|\psi_1 \rangle = \dfrac{|00\rangle + |10\rangle}{\sqrt{2} } $, and $|\psi_2 \rangle = |S'\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2} } $.

Now, if you apply $\big( H \otimes I \big) \cdot CNOT $ to the state $|S'\rangle$ then this is equivalent to applying $U_1 = CNOT \cdot \big(H\otimes I\big) \cdot \big( H \otimes I \big) \cdot CNOT $ to $|S\rangle = |00\rangle$ since
\begin{align}
\bigg( \big( H \otimes I \big) \cdot CNOT \bigg) |S'\rangle &= \bigg( \big( H \otimes I \big) \cdot CNOT \bigg) \bigg( CNOT \cdot \big( H\otimes I\big) |00\rangle \bigg) \\
&= \big( H \otimes I \big) \cdot CNOT \cdot CNOT \cdot \big(H\otimes I\big) |00\rangle \\
&= \big( H \otimes I \big) \cdot \big(H\otimes I\big) |00\rangle \\
&= |00\rangle
\end{align}

Note: $CNOT \cdot CNOT = I$ since $CNOT$ is its own inverse. Similarly, $H \otimes I$ is its own inverse hence $\big( H \otimes I \big) \cdot \big( H \otimes I \big) = I$.

You can see this on the quantum circuit as follows:

1Thank you very much for taking the time to write such detailed explanation! I really appreciate it. I'm learning these on my own and generous people like you make our lives easier! thank you – user206904 – 2021-01-01T15:13:34.213