In the following I shall assume that $f$ is continuous with compact support.

It is known that if $\varphi : \mathbb R^n \to \mathbb R$ has $\nabla \varphi \ne 0$ at all the points of $H = \varphi^{-1} (0)$, then
$$\int _{\mathbb R^n} f(v) \, \delta (\varphi(v)) \, \mathrm d v = \int _H \frac{f(s)} {\|\nabla \varphi(s)\|} \, \mathrm d s \ ,$$
the latter integral being with respect to the natural measure induced on $H$.

In your case $\varphi(x) = \langle w, x \rangle - b$, so $\| \nabla \varphi \| = \| w \| = 1$.

Since $w \ne 0$, let us assume that $w_n \ne 0$. Consider the (global) parametrization $h : \mathbb R^{n-1} \to H_{w,b}$ given by
$$h(u_1, \dots, u_{n-1}) = \left( u_1, \dots, u_{n-1}, \frac 1 {w_n} (b - u_1 \, w_1 - \dots - u_{n-1} \, w_{n-1}) \right) .$$

In this paramerization, the coefficients of the Riemannian metric on $H_{w,b}$ are $g_{ii} = 1 + \frac {w_i ^2} {w_n ^2}$ and $g_{ij} = \frac {w_i w_j} {w_n ^2}$ for $i < j$, so your integral becomes
$$\sqrt {\det (g_{ij})} \int _{\mathbb R ^{n-1}} f \left( u_1, \dots, u_{n-1}, \frac 1 {w_n} (b - u_1 \, w_1 - \dots - u_{n-1} \, w_{n-1}) \right) \ \mathrm d u_1 \dots \mathrm d u_{n-1} \ .$$

Since $f$ is continuous with compact support, the integral written above exists and is finite. Using Lebesgue's dominated convergence theorem, if the sequence $w^{(k)}$ converges to $w$ then you also get the convergence of the corresponding integrals.

The determinant, too, is continuous on some neighbourhood of some given $w$: if the coordinate $w_n$ is non-zero, then it is so on some neighbourhood $U$ of $w$ in $S^{n-1}$ (by the continuity of the coordinate functions), so you may safely divide by it; since all the operations involved in that determinant al algebraic, the determinant will depend continuously on $w$ in the neighbourhood $U$. Since continuity is a local property, this is enough.

To conclude: $f$ being continuous with compact support is enough for the continuity of its Radon transform (with respect to $w$ and $b$ jointly); it is up to you to decide whether this is too restrictive or not. This condition (continuity and compact support) is sufficient to ensure that the restriction of $f$ to any hyperplane $H_{w,b}$ is integrable. Absent it, I do not know what to replace it with.