The Radon transform for $f\in C_c^\infty(\mathbb{R}^n)$ is defined by
\begin{equation} Rf(\theta , s) := \int _{\{ x^T\theta =s \}} f(x)dx. \ \ \ (1) \end{equation}
Now it says, that an equivalent representation is given by
\begin{equation} Rf(\theta , s) = \int _{\mathbb{R}^n}f(x)\delta(s-x^T\theta)dx, \ \ \ (2) \end{equation}
where $\delta$ is the $\delta$-distribution with the symbolic notation
\begin{equation} f(a)= \int_{\mathbb{R}^n}f(x)\delta(x-a)dx, \ \ \ (3) \end{equation}
but I don't see why. In (2), $\delta$ seems to be a function of a scalar-valued argument, whereas in (3) the argument is a vector, so we can't really use this to rewrite (2) and even if so, then (2) would only give us the evaluation of $f$ at a certain point, not (1). Probably I'm missing something simple right now, but I can't figure it out. Thank you for any hints.
This follows directly from the definition of the Radon function in $\mathbb R^n$.
The Radon transform $Rf$ of a function $f:\mathbb R^2 \to \mathbb R$ is a function defined on the space of straight lines in $\mathbb R^2$ by the line integral along each such line,
$$Rf: [0, 2\pi]\times\mathbb R \to \mathbb R$$ $$Rf(\alpha, s)= \int_{-\infty}^{\infty}f(s\alpha + z\alpha^{\perp})dz$$
where $\alpha=(\cos \alpha,\sin \alpha)$ is the normal vector of the line and $\alpha^{\perp}$ is the tangent vector.
More generally, the Radon transform $Rf$ of a function $f:\mathbb R^n \to \mathbb R$ is a function defined on the space of all hyperplanes in $\mathbb R^n$. If one parametrizes these hyperplanes by $\{x\in\mathbb R^n : x \cdot \alpha=s\}$ where $\alpha \in S^{n-1}$ is a unit vector of $\mathbb R^n$ and $s\in\mathbb R$, one obtains a function defined on $S^{n-1}\times\mathbb R$ by
$$Rf: S^{n-1}\times\mathbb R \to \mathbb R$$ $$Rf(\alpha, s)= \int_{x \cdot \alpha=s}f(x)dx=\int_{\alpha^{\perp}}f(s\alpha + y)dy$$
where $\alpha$ represents angles: one angle in 2D (points on the unit circle or equivalently tangent lines to the unit circle), two angles in 3D (points on the unit sphere or equivalently tangent planes to the unit sphere), and so on. The number $s$ is the (signed) distance between these hyperplanes and the origin.
Notice that this is equivalent to your definition of the Radon transform as
$$ Rf(\theta , s) = \int _{\{ x^T\theta =s \}} f(x)dx=\int_{x \cdot \theta=s}f(x)dx =\int_{\theta^{\perp}}f(s\theta + y)dy $$
which we could now compare to
$$Rf(\theta , s) = \int _{\mathbb{R}^n}f(x)\delta(s-x^T\theta)dx=\int _{\mathbb{R}^n}f(x)\delta(s-x \cdot \theta)dx$$
As $f\in C_c^\infty(\mathbb{R}^n)$, we know from a property of convolutions that
$$f(x)=\int_{\mathbb{R}^n}\delta(x-y)f(y)dy$$
and since we have the constraint that ($x^T\theta =s$) or ($x \cdot \theta=s$) we see that the convolution property allows us to write
$$Rf(\theta , s) = \int _{\mathbb{R}^n}f(x)\delta(s-x^T\theta)dx=\int _{\mathbb{R}^n}f(x)\delta(s-x \cdot \theta)dx = \int_{\theta^{\perp}}f(s\theta + y)dy = \int _{\{ x^T\theta =s \}} f(x)dx$$