Here $|\psi\rangle$ is an $n$ component unit vector with $|x\rangle$ being one of the unit basis vectors, i.e., $|\psi\rangle=a_1|x_1\rangle+a_2|x_2\rangle+\cdots+a_x|x\rangle+\cdots+a_n|x_n\rangle$ and $\exp(-iHt)$ is a unitary operator since $H=|x\rangle\langle x|+|\psi\rangle\langle \psi|$ is hermitian.
$$ H=|x\rangle\langle x|+|\psi\rangle\langle \psi|\quad\&\quad\exp(-iHt)\approx I-itH\\ \exp(-iHt)|\psi\rangle=(I-itH)|\psi\rangle=(I-it|x\rangle\langle x|-it|\psi\rangle\langle \psi|)|\psi\rangle\\ =(1-it)|\psi\rangle-it\langle x|\psi\rangle|x\rangle $$
How can we interpret from $|\psi(t)\rangle=\exp(-iHt)|\psi\rangle=(1-it)|\psi\rangle-it\langle x|\psi\rangle|x\rangle$ that the vector $|\psi\rangle$ is rotated into the $|x\rangle$ direction by the action of $\exp(-iHt)$ ?
Reference : Page 257, Quantum Computation and Quantum Information by Nielsen and Chuang
My Attempt
$$ |\psi\rangle=a_1|x_1\rangle+a_2|x_2\rangle+\cdots+a_x|x\rangle+\cdots+a_n|x_n\rangle\\ \langle x|\psi\rangle=a_x\\ \langle x|\exp(-iHt)\psi\rangle=(1-it)a_x-ita_x=(1-2it)a_x=\frac{1-2it}{\sqrt{1+4t^2}}\times\sqrt{1+4t^2}\times a_x=e^{i\theta}\sqrt{1+4t^2}\times a_x $$ Consider another state $e^{-i\theta}|\psi\rangle$ which has the same characteristics as the state $|\psi\rangle$ because having an overall phase does not change the probability which is the amplitude squared.
The angle between $e^{-i\theta}|\psi\rangle$ and $|x\rangle$ is such that $\cos\theta=\sqrt{1+4t^2}a_x>a_x\implies$ the angle is lesser than that between the original state $|\psi\rangle$ and $|x\rangle$.
Is it an explanation for the required statement in my reference?
You've left out or missed a very important piece of information: the Hamiltonian is defined as
$$H=|x\rangle\langle x|+|\psi\rangle\langle\psi|$$
They then truncate the expansion of the exponential Hamiltonian:
$$\exp(-i Ht)=\sum_{n=0}^\infty \frac{(-i)^n H^n t^n}{n!}=I-itH-\frac{H^2 t^2}{2}+...\approx I-itH$$
Then
$$\exp(-i Ht)|\psi\rangle\approx (I-itH)|\psi\rangle$$
Using this truncation and the definition of $H$ gives
$$(I-itH)|\psi\rangle=(I-it|x\rangle\langle x|-it|\psi\rangle\langle\psi|)|\psi\rangle=(1-it)|\psi\rangle-it \langle x|\psi\rangle|\psi\rangle$$
EDIT: So when they say its been rotated slightly in the $|x\rangle$ direction they mean
$$|\langle x|\psi\rangle|^2<|\langle x|\exp(-iHt)|\psi\rangle|^2$$
To see this, note that
$$\langle x|\exp(-iHt)|\psi\rangle\approx \langle x|[(1-it)|\psi\rangle-it \langle x|\psi\rangle|\psi\rangle]$$
$$=\langle x|\psi\rangle-it\langle x|\psi\rangle-it\langle x|\psi\rangle^2=a_x-it a_x-it a_x^2$$
However, note that we can multiply $|\psi\rangle$ by $e^{i\phi}$ for any $\phi\in\mathbb{R}$ without changing the overall absolute value. Thus, choose $\phi$ such that
$$a_x\in\mathbb{R}$$
Taking the absolute value of this (with $a_x=a$) gives
$$|a_x-it a_x-it a_x^2|^2=a^2 (1 + (1 + a)^2 t^2 )$$
while on the other hand we have
$$|\langle x|\psi\rangle|^2=|a_x|^2=a^2$$
Thus, to show we have rotated in the $|x\rangle$ direction we simply need to show
$$a^2 (1 + (1 + a)^2 t^2 )-a^2 >0$$
This is equivalent to showing that
$$ (1 + a)^2 t^2 >0$$
which is always true unless $|\langle x|\psi\rangle|=1$. But this is fine since that would imply $|\psi\rangle$ was already pointed in the $|x\rangle$ direction so $H$ would have no effect on it.