Say I have a wave equation of the form $$\nabla^{2}f(t,\mathbf{x})=\frac{1}{v^{2}}\frac{\partial^{2}f(t,\mathbf{x})}{\partial t^{2}}$$ which is clearly a partial differential equation (PDE) in $\mathbf{x}$ and $t$.
Is it valid to consider a solution in which one Fourier decomposes the spatial part of the function $f(t,\mathbf{x})$ and not its temporal part. That is, is it ok to express the solution in the form $$f(t,\mathbf{x})=\int_{-\infty}^{+\infty}\frac{d^{3}k}{(2\pi)^{3}}\tilde{f}(t,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}}$$ And if so, why is it ok to do this?
[Is it simply that one Fourier decomposes the solution at a particular fixed instant in time, $t$ and then requires that the solution has this form for all $t$, hence the mode functions $\tilde{f}(t,\mathbf{k})$ must satisfy the PDE $$\frac{\partial^{2}\tilde{f}(t,\mathbf{k})}{\partial t^{2}}+v^{2}\mathbf{k}^{2}\tilde{f}(t,\mathbf{k})=0$$ or is there some other reasoning behind it?]
When considering a function $f(t,\mathbf x)$ on $[0,\infty)\times \mathbb{R}^n$, we can focus on one time slice at a time, fixing $t$ and dealing with a function of $\mathbf x$ only. The Fourier transform can be applied to this slice, since it's just a function on $\mathbb{R}^n$. The result can be denoted $\tilde f(t,\mathbf k)$, and is usually called the Fourier transform with respect to spatial variable (or sometimes, "partial Fourier transform"). As long as $t$ is fixed, nothing new happens: for example, $$f(t,\mathbf{x})=\int_{\mathbb{R}^n}\frac{d\mathbf k}{(2\pi)^{3}}\tilde{f}(t,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}}\tag{1}$$ is just the inversion formula for Fourier transform.
One eventually wants to understand how the solution evolves in time, so the time derivative has to be taken. Formally speaking, we differentiate (1) under the integral sign, which of course this needs a justification, as always when this trick is performed. The lecture notes Using the Fourier Transform to Solve PDEs by Joel Feldman present such calculations for the wave equation.