$\nabla^{2}\mathbf{E}-\frac{1}{c^{2}}\mathbf{E}_{tt}=0 \tag{1}$
is equivalent to Helmhotz equation by the Fourier transformation i.e.
$\tilde{E}_{zz}(z,w)+\epsilon(w)\frac{w^{2}}{c^{2}}\tilde{E}(z,w)=0 \tag{2} $
where
$$\tilde{E}(z,w)=\int_{-\infty}^{\infty}{E}(z,t)e^{\dot\imath wt}\mathrm{d}t$$
and
$$\epsilon(w)=1+\tilde{\chi}^{(1)}(w)=1+\int_{-\infty}^{\infty}{\chi^{(1)}}(t)e^{\dot\imath wt}\mathrm{d}t.$$
I am facing a problem to reduce the equation (1) to equation (2) by Fourier transform.
It seems to me that this is a rather physicsy question. To make things clear, if we talk about susceptibility $\chi$, we are talking about electrodynamics in media. Here, we assume that we are dealing with a spatially homogeneous, isotropic and linear but dispersive medium, such that susceptibility is space-independent but frequency dependent. These are necessary for obtaining the result as we will see and therefore it seems to me at least that the equivalence you propose between the wave equation and the dispersive Helmholtz equation is not correct.
The first problem is that the $c$ in your original equation is not actually the full speed of light of vacuum $c_o=1/\sqrt{\epsilon_o \mu_o}$, but rather $c^2= c_o^2/(1+\chi)$ (see eqns. (2.5.8-11) of these notes). Hence, we are actually considering the dispersive wave equation
$$\nabla^2 \mathbf{E}(\mathbf{x},t) - \frac{(1+\chi)}{c_o^2}\partial_t^2\mathbf{E}(\mathbf{x},t) =0, \tag{1} $$ where $\mathbf{E}(\mathbf{x},t)$ is the electric field and $\mathbf{x}=(x,y,z)$ is position.
Now, in principle, we would like to simply take the Fourier transform of this equation. To find the relation you stated, we will however first talk about susceptibility. First remember that by the assumed linearity of our medium the polarisation density $\mathbf{P}(\mathbf{r},t)$ and the electric field $\mathbf{E}(\mathbf{x},t)$ are related by $\mathbf{P}(\mathbf{r},t) = \epsilon_o\chi \mathbf{E}(\mathbf{x},t)$, eqn. (5.2-1) of the notes. In a linear dispersive medium, the electric field plays the role of the source of polarisation in some linear ODE in time, $L \mathbf{P} = \mathbf{E}$ (e.g. $L=\alpha_2\partial_t^2 + \alpha_1\partial_t + \alpha_o)$). This equation may be solved for $\mathbf{P}$ in terms of the Greens function $\epsilon_o G(t-t')$ of $L$ to yield $$ \mathbf{P}(\mathbf{x},t) =\epsilon_o \int G(t-t') \mathbf{E}(\mathbf{x},t') \mathrm{d}t'. $$ If both $\mathbf{P}$ and $\mathbf{E}$ are Fourier expanded in time, we find, after renaming $u=t-t'$, $$ \int \tilde{\mathbf{P}}(\mathbf{x},\omega) e^{-i\omega t}\mathrm{d} \omega = \epsilon_o \int \underbrace{\left( \int G(u) e^{i\omega u} \mathrm{d}u\right)}_{\equiv \chi(\omega)} \tilde{\mathbf{E}}(\mathbf{x},\omega) e^{-i\omega t} \mathrm{d}\omega $$ By comparing coefficients, $ \tilde{\mathbf{P}} = \epsilon_o\chi(\omega) \tilde{\mathbf{E}}$. Hence, for a wave of frequency $\omega$, the susceptibility is $\chi(\omega)$ as defined above.
Now we can take the temporal Fourier transform of the wave equation (1), to find the result you wanted. Since it is a linear equation, this corresponds to substituting a monochromatic wave for $\mathbf{E}$ i.e. $\mathbf{E}(\mathbf{x},t)= \tilde{\mathbf{E}}(\mathbf{x}) e^{i\omega t} $. Taking the derivatives of this expression and cancelling the exponential factor, we obtain: $$ \nabla^2 \tilde{\mathbf{E}}(\mathbf{x},\omega) + \frac{\omega^2(1+\chi(\omega))}{c_o^2}\tilde{\mathbf{E}}(\mathbf{x},\omega) =0. $$
This is the equation you were looking for. Don't hesitate to ask if anything is unclear.