Generalizing a formula with distributions — Distributional Radon transform I will try to describe the problem, it will necessarily be incomplete, so please if you have questions or remarks to make it more clear do not hesitate to leave them in comments.
The problem
Let $a$ be a constant, we know that a PDE (let us call it ‘the equation’) has solutions of the form
$$u(x)= \int_{S^2}\alpha(x\cdot\omega+ a,\ \omega)\,d\sigma(\omega)\quad , \ x\in \mathbb{R}^3$$
where
\begin{align*}
    \alpha(s,\omega)&= \frac{\partial}{\partial s}\mathscr{R}g(s,\omega)+ \frac{\partial^2}{\partial s^2}\mathscr{R}f(s,\omega) 
\end{align*}
for given functions $f,g \in C_0(\mathbb{R}^3)$. 
And $\mathscr{R}:C_0(\mathbb{R}^3)\rightarrow C_0(\mathbb{R}\times S^2)$ is the Radon transform in the $3$ dimensional space.
The question is, how would we generalize $u$ to a distribution so that we include the cases when $\mathscr{R}f$ and $\mathscr{R}g$ are not differentiable (with respect to $s$)? 
It is clear that $\alpha$ can be regarded as a distribution in $s$.
The trial
Here is what we can try for example, we want to generalize $u(x)$ to a distribution $U\in S'(\mathbb{R}^3)$, which is defined by
\begin{align*}
    U(\phi)&= \int_{\mathbb{R}^3} u(x)\phi(x) \,dx \\
    &= \int_{\mathbb{R}^3} \int_{S^2}\alpha(x\cdot\omega+a,\omega)\phi(x) \,d\sigma(\omega)\,dx \quad \text{ for all } \phi\in S(\mathbb{R}^3).
\end{align*}
To each $\omega\in S^2$ correspond two vectors $\nu, \eta\in S^2$ (infinitely many choices) such that $(\omega, \nu, \eta)$ is an orthonormal basis of $\mathbb{R}^3$, we have the following bijection (the change of coordinates) as an isometry;
\begin{align*}
    \theta_\omega &: \ \mathbb{R}^3\rightarrow \mathbb{R}^3 \\
    & \ \ \ x\mapsto (x\cdot\omega,\ x\cdot\nu,\ x\cdot\eta)^t
\end{align*}
Letting $s= x\cdot\omega+a$, $y_1= x\cdot\nu$ and $y_2=x\cdot\eta$, we can write
$$x= (x\cdot\omega)\omega+ y_1\nu+ y_2\eta= s\omega+ y_1\nu+ y_2\eta- a\omega $$
then for $y= (y_1, y_2)^t\in \mathbb{R}^2$ we can define the test function $\tau_y\in S(\mathbb{R})$ by $\tau_y(s)= \phi(x)$.
And we can then consider a distribution induced by $\alpha(s,\omega)$ (as a function of $s$) that we can denote $\alpha_\omega$:
\begin{align*}
    U(\phi)&= \int_{\mathbb{R}^3} \int_{S^2}\alpha(s,\omega)\tau_y(s) \,d\sigma(\omega)\,dx \quad \text{ by the change of variable we want to apply } \\
    &= \int_{S^2} \int_{\mathbb{R}^2}\left(\int_{\mathbb{R}}\alpha(s,\omega)\tau_y(s) \,ds\right)\,dy \,d\sigma(\omega) \quad \text{ isolating $s$ before integrating over $\mathbb{R}^2$ }\\
    &= \int_{\mathbb{R}^2}\int_{S^2} \alpha_\omega(\tau_y) \,d\sigma(\omega)\,dy \quad \quad\quad \text{ here $\alpha_\omega$ is a distribution acting on $\tau_y$.}
\end{align*}
Let us assume we will find no trouble in considering that integrating the distribution over $S^2$ then over $\mathbb{R}^2$ is well defined.
Now for distributions $f,g\in S'(\mathbb{R}^3)$ of compact support, and $\psi\in S(\mathbb{R}^2)$ such that $\psi(y)= 1$, $\forall y\in P_{\mathbb{R}^2}(B_r(0)) $ (the projection of the ball containing the compact support of $f$ and $g$ on the subspace $\mathbb{R}^2$).
We want to define a ‘distributional Radon transform’ $\mathscr{R}_\omega h$ (of any function $h$):
$$\mathscr{R}_\omega h \in S'(\mathbb{R}) : \phi\mapsto h\big( (\phi\otimes\psi) \circ\theta_\omega\big) $$
where the tensor product of the two test functions is defined by $\phi\otimes\psi\begin{pmatrix}
s \\
y_1 \\
y_2
\end{pmatrix}= \phi(s)\psi(y)$ for all $(s, y_1, y_2)^t\in \mathbb{R}^3$.
And then it will follow that $$\alpha_\omega= \mathscr{R}_\omega g '+\mathscr{R}_\omega f''.$$
Or $$\alpha_\omega(\tau)= -\mathscr{R}_\omega g (\tau')+ \mathscr{R}_\omega f(\tau'') \quad \forall \tau \in S(\mathbb{R}).$$
This last formula is then used to retrieve the distribution $U$.
The question
Is any of that making any sense? 
The classical Radon transform $\mathscr{R}h$ captures the sum of the density $h$ over some Euclidean plane $P_{(s,\omega)}$, whereas this distributional Radon transform ‘mimics’ this but for distributions $h$!
Would you think of that problem in another way?
 A: Edit. While my first answer was more a suggestion on how to proceed, I decided to expand it in full. I also changed the former notation at some point in order to make it clearer, while still retaining the "duality bracket" notation for distributions instead of the "functional" one since using the latter would probably led to somewhat cumbersome formulas.

*

*The first thing to note is that the generalization of the Radon transform to slowly increasing  distributions and to fully general distributions uses the concept of adjoint (Radon) transform (see D. Ludwig, [1], or S. R. Deans, [2], chapter 5, pp. 109-124) (§5.4, pp. 120-130).
First of all let's define the integral operator $\mathscr{R}^\dagger$  as
$$\DeclareMathOperator{\dmu}{d\!}
\mathscr{R}^\dagger \hat{\psi}(\mathbf{x})\triangleq  \int\limits_{\lvert\xi\rvert=1} \hat{\psi}(\xi\cdot\mathbf{x},\xi)\dmu\xi\quad\mathbf{x}\in\Bbb R^{n}\label{1}\tag{AR}
$$
where

*

*$\hat{\psi}(p,\xi)=\mathscr{R}\psi (p,\xi)$ with $p\in\Bbb R$ is the Radon transform of a $C^\infty$ function $\psi$ and

*$\xi\in\Bbb S^{n-1}=\big\{\xi\in\Bbb R^n:\lvert\xi\rvert=1\big\}$.
This notation will be used throughout below: $\psi$  and likewise the all the (test) functions used below may be readily assumed to be rapidly decreasing, compactly supported, etc.: however, to keep the answer as elementary as possible, it is only assumed that their behavior at infinity is such that all the integration shown are meaningful (as it is the case for functions in $\mathscr{S}$ and $\mathscr{D}$). Then, for any two given and properly behaved $C^\infty$ functions $q$, $r$ we have that
$$
\begin{split}
\langle q, \mathscr{R}^\dagger \hat{r}\rangle & = \int\limits_{\Bbb R^n} q(\mathbf{x}){\mathscr{R}^\dagger\hat{r}}(\mathbf{x})\dmu\mathbf{x} \\
& = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}{\hat{r}}(\xi\cdot\mathbf{x},\xi)\dmu\xi\bigg]\dmu\mathbf{x}\\
& = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}\bigg(\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\delta(p-\xi\cdot\mathbf{x})\dmu p\bigg)\dmu\xi\bigg]\dmu\mathbf{x}\\
& =  \int\limits_{\lvert\xi\rvert=1}\bigg[\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\bigg( \int\limits_{\Bbb R^n} q(\mathbf{x})\delta(p-\xi\cdot\mathbf{x})\dmu\mathbf{x}\bigg)\dmu p\bigg]\dmu\xi \\
& =  \int\limits_{\lvert\xi\rvert=1}\,\,\int\limits_{-\infty}^{+\infty}{\hat{q}}(p,\xi)\hat{r}(p,\xi)\dmu p\dmu\xi =\langle \mathscr{R}{q},\hat{r}\rangle=\langle \hat{q},\hat{r}\rangle.
\end{split}\label{2}\tag{GR}
$$
Since the first member on the right of \eqref{2} can be readily interpreted as a lowly decreasing or general distribution, it is assumed as the definition of the Radon transform of a distribution, be it in $\mathscr{S}^\prime$ or in $\mathscr{D}^\prime$.


*The second thing to note is that $\mathscr{R}^\dagger$ is verbatim the adjoint transform of the Radon transform, thus we have trivially that $$
\langle \mathscr{R}^\dagger \hat{q}, r\rangle\triangleq\langle \hat{q}, \mathscr{R} r \rangle=\langle \hat{q},\hat{r}\rangle.
$$
This, as it will be shown in the next point, is important in order to understand the operational structure of $u$ as a distribution.


*The third thing to note is that the function $u(\mathbf{x})$ in the OP we are trying to understand as a distribution is simply a translation of amplitude $a\in\Bbb R$ of the adjoint transform \eqref{1} of the Radon transform of some function. Precisely, still assuming that we are dealing with functions, we see that
$$\DeclareMathOperator{\dmu}{d\!}
u(\mathbf{x})= \int\limits_{\lvert\xi\rvert=1} {\alpha}(\xi\cdot\mathbf{x}+a,\xi)\dmu\xi = \mathscr{R}^\dagger {\alpha}_a(\mathbf{x}) \quad\mathbf{x}\in\Bbb R^{n}\label{3}\tag{AR'}
$$
where we define
$$
\alpha_a(p,\xi)\triangleq \alpha(p+a,\xi)\quad \forall a\in\Bbb R.
$$


*Finally, using the standard definitions of derivative and translation of a distribution (as for example given in [3]), we have all the elements in order to express $u$ as a general distribution: precisely.
$$
\begin{split}
{u}(\varphi) & = {\mathscr{R}^\dagger \alpha_a}(\varphi) = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\mathscr{R}g_a+ \tfrac{\partial^2}{\partial p^2}\mathscr{R}f_a\right]}(\varphi) \\
& = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\hat{g}_a+ \tfrac{\partial^2}{\partial p^2}\hat{f}_a\right]}(\varphi)\\
& = {\mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{g}_a(\varphi) + \mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{f}_a}(\varphi) \\
&= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\mathscr{R}\varphi\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\mathscr{R}\varphi\Big\rangle \\
&= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\hat{\varphi}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\hat{\varphi}\Big\rangle\\
&= \Big\langle\tfrac{\partial}{\partial p}\hat{g},\hat{\varphi}_{-a}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f},\hat{\varphi}_{-a}\Big\rangle\\
&= - \Big\langle\hat{g},\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle\hat{f},\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle\\
&= - \Big\langle g, \mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle f,\mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle
\end{split}
$$
References
[1] Stanley Roderick Deans, The Radon transform and some of its applications, revised reprint of the 1983 original edition (English), Malabar, FL: Publishing Co., ISBN 0-89464-718-0, pp. xii+295 (1993), MR1274701, Zbl 0868.44001.
[2] Donald Ludwig, "The Radon transform on Euclidean space", (English)
Communication in Pure and Applied Mathematics 19, 49-81 (1966), MR0190652, Zbl 0134.11305.
[3] Vasily Sergeevich Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
