Think of a vector $\mathbf{V}$. As seen in a coordinate system $S$ with basis vectors $\hat{e}_i$, it can be written
$$\mathbf{V} = \sum_i V_i \hat{e}_i$$
where $V_i$ are the components of $\mathbf{V}$ in $S$. As seen from another coordinate system $S'$ with basis vectors $\hat{e}_i'$, it has a representation
$$\mathbf{V} = \sum_i V_i' \hat{e}_i'.$$
Obviously the length of the vector is independent of the coordinate system used to represent it. In other words, we must have
$$\sum_i V_i^2 = \sum_i (V_i')^2$$
Proceeding with this analogy, for a function $f(x)$ one can have a position space representation in $\delta$-function basis as
$$f(x) = \int f(x') \delta(x-x') dx'$$
where the "component" of $f(x)$ along the "basis vector" $\delta(x-x')$ is $f(x')$ and we sum (integrate since $x$ is a continuous variable) over all the possible "axes".
One can look at the same function in Fourier-space representation as
$$f(x) = \int g(k) e^{-i k x} dk$$
where $e^{-ikx}$ are the "basis vectors" and $g(k)$ are the "components" of $f(x)$ along these basis vectors.
You would then agree that
$$ \int |f(x)|^2 dx = \int |g(k)|^2 dk$$
So Parseval's theorem is just the restatement of the invariance of the length of a "vector" independent of the representation used.
If $|f(x)|^2$ is proportional to the energy, then Parseval's theorem is a statement of the conservation of the energy as seen in the real-space domain or the Fourier-space domain
If $f(x)$ is a quantum-mechanical wavefunction, $|f(x)|^2$ is proportional to the probability density. Parseval's theorem is then a statement of the conservation of the probability as seen in the position-space representation or the momentum-space representation.
See also Parseval's identity