On pg. 46 of Principles of Quantum Mechanics:
Question: What is going on with the red highlighted portion of (1.8.34)?
It seems like (1.8.34) follows completely from (1.8.33) (i.e. subtracting one side from the other). Isn't this true? So How does the operator $\frac{d^2}{dt^2}-\Omega$ have anything to do with deriving (1.8.34)?
EDIT: The proof later goes on to state that
How does this follow?
Regarding the first question: begin with (1.8.32), $$ |x(t) \rangle = |\text{I}\rangle x_\text{I}(t) + |\text{II}\rangle x_{\text{II}}(t). $$ Applying $(\frac{d^2}{dt^2} - \Omega)$ to the left side gives us $$ (\frac{d^2}{dt^2} - \Omega)|x(t) \rangle = |\ddot x(t) \rangle - \Omega |x(t)\rangle = |0 \rangle. $$ Applying the operator to the right side gives us \begin{align} \left(\frac{d^2}{dt^2} - \Omega \right)[|\text{I}\rangle x_\text{I}(t) &+ |\text{II}\rangle x_{\text{II}}(t)] = \left(\frac{d^2}{dt^2} - \Omega \right)|\text{I}\rangle x_I(t) + \left(\frac{d^2}{dt^2} - \Omega \right)|\text{II}\rangle x_{\text{II}}(t) \\ & = |\text{I}\rangle \ddot x_{\text{I}}(t) - \Omega |\text{I}\rangle x_\text{I}(t) + |\text{II}\rangle \ddot x_{\text{II}}(t) - \Omega |I\rangle x_{\text{II}}(t) \\ & = |I\rangle \ddot x_{I}(t) - - \omega_I^2|I\rangle x_I(t) + |\text{II}\rangle \ddot x_{\text{II}}(t) - - \omega_{\text{II}}^2 |I\rangle x_{\text{II}}(t) \\ & = |\text{I}\rangle (\ddot x_\text{I}(t) + \omega_\text{I}^2 x_\text{I}(t)) + |\text{II}\rangle (\ddot x_{\text{II}}(t) + \omega_{II}^2 x_{\text{II}}(t)), \end{align} which indeed gives us the right-hand side of (1.8.34).
Regarding the second question: (1.8.31a) tells us that the "ket" $|\text{I}\rangle$ corresponds to $\frac 1{\sqrt{2}}\left( \begin{smallmatrix} 1\\1 \end{smallmatrix}\right)$ in the $1,2$ basis. To get the corresponding "bra", take the conjugate-transpose: $$ \langle \text{I} | \leftrightarrow \frac 1{\sqrt{2}}\pmatrix{1 & 1}. $$ Plugging that in yields $$ \langle \text{I} |x(0) \rangle = \frac 1{\sqrt{2}} \pmatrix{1 & 1} \pmatrix{x_1(0)\\x_2(0)}, $$ and matrix multiplication yields $\frac{x_1(0) + x_2(0)}{\sqrt{2}}$. The second line can be derived similarly.