Let $\Omega\subset\mathbb{R}^2$ be a bounded domain with a $C^1$ boundary. Let $f$ be zero outside of $\Omega$ and $C^2$ inside, with $f=0$ on $\partial\Omega$. Also assume that all of the partial derivatives of $f$ extend continuously to $\partial\Omega$. Let us define
$$
g(x) = \begin{cases}
\Delta f(x) &\text{if $x\in\Omega$},\\\\
0 &\text{otherwise}.
\end{cases}
$$
Now, $g$ is not equal to $\Delta f$ on all of $\mathbb{R}^2$. In fact, $\Delta f$ may not even exist as a function on $\mathbb{R}^2$. But it does exist as a distribution. Let us see how $\Delta f$, the distribution, compares with $g$.
Let $\varphi$ be a smooth test function. Then
$$
\langle\Delta f,\varphi\rangle = \langle f,\Delta\varphi\rangle
= \int_{\mathbb{R}^2}f(x)\Delta\varphi(x)\,dx.
$$
Since $f=0$ outside of $\Omega$, we need only integrate over $\Omega$. We can then use Green's second identity to obtain
$$
\int_\Omega f(x)\Delta\varphi(x)\,dx
= \int_\Omega g(x)\varphi(x)\,dx
+ \int_{\partial\Omega}
(f\partial_{\bf n}\varphi - \varphi\partial_{\bf n}f)\,d\sigma,
$$
where $\partial_{\bf n}$ is the derivative in the direction of the outward normal, and $\sigma$ is the surface measure on $\partial\Omega$. (Since we are in $\mathbb{R}^2$, this is just the arclength measure.) Since $f=0$ on $\partial\Omega$, we get
\begin{align*}
\langle\Delta f,\varphi\rangle
&= \int_\Omega g\varphi\\,dx
-\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\\,d\sigma\\\\
&= \int_{\mathbb{R}^2} g\varphi\\,dx
-\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\\,d\sigma\\\\
&= \langle g,\varphi\rangle
-\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\\,d\sigma.
\end{align*}
What this shows is that $\Delta f = g + \nu$, where $\nu$ is the distribution that maps $\varphi$ to
$$
\langle\nu,\varphi\rangle
= -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\,d\sigma.
$$
For example, if $\partial_{\bf n}f(x)=-1$ for all $x\in\partial\Omega$, then $\nu$ is just the surface measure on $\partial\Omega$. This is the case in the example I gave in my answer to the other question. And in that case, the surface measure on the boundary of $[0,\pi]$ in $\mathbb{R}$ is just two point masses, one at $0$ and one at $\pi$.
Lastly, to address a comment from the other question, the Laplacian is still rotationally invariant when interpreted as a distributional derivative. To prove this, we apply $\Delta f$ to a smooth test function, then move the Laplacian over to the test function (where it acts in the classical way), and then utilize the rotational invariance of the classical Laplacian.
A fairly short and accessible reference for tempered distributions, which is free online, is Chapter 11 of Applied Analysis by Hunter and Nachtergaele. Also, there is a chapter on the Laplace operator in Folland.
Edit:
To address the question in the comments, the Radon transform of a tempered distribution is defined by $\langle Rf,\varphi\rangle=\langle f,R^*\varphi\rangle$. More precisely, it is $\langle Rf,\varphi\rangle_{S^1\times\mathbb{R}}=\langle f,R^*\varphi\rangle_{\mathbb{R}^2}$. The inner product on the right is the usual $L^2$ inner product on $\mathbb{R}^2$; the inner product on the left is defined by
$$
\langle f,g\rangle_{S^1\times\mathbb{R}}
= \frac1{2\pi}\int_0^{2\pi}\int_{\mathbb{R}}
f(\theta,s)g(\theta,s)\,ds\,d\theta.
$$
It follows that
\begin{align*}
\langle R(\Delta f),\varphi\rangle
&= \langle \Delta f,R^\*\varphi\rangle
= \langle f,\Delta(R^\*\varphi)\rangle\\\\
&= \langle f,R^\*(\partial_s^2\varphi)\rangle
= \langle Rf,\partial_s^2\varphi\rangle
= \langle \partial_s^2(Rf),\varphi\rangle,
\end{align*}
and so $R(\Delta f)=\partial_s^2(Rf)$, even in the distributional sense.
What is relevant for your other question is how to compute $R\nu$. My guess is that you are asking if
$$
R\nu = -\sum_{x\in\partial\Omega\cap L}\partial_{\bf n}f(x).
$$
The intuition behind this formula does not work, because it does not account for the angle between $L$ and $\partial\Omega$ at the point of intersection. I will leave it as an exercise to show that if $\sigma$ is the arclength measure on $S^1$, that is, if
$$
\langle\sigma,\varphi\rangle = \int_0^{2\pi}\varphi(e^{i\theta})\,d\theta,
$$
then
$$
R\sigma(\theta,s) = 2\frac1{\sqrt{1 - s^2}}\chi_{[-1,1]}(s).
$$
The $2$ comes from the two points where the line intersects the circle, and the factor of $|1-s^2|^{-1/2}$ comes from the angle of intersection. If $L$ is the line corresponding to $(\theta,s)$ and $\alpha(L)\in[0,\pi/2]$ is the angle with which $L$ intersects $S^1$, then $|s|=\cos\alpha(L)$. We therefore have
$$
R\sigma(\theta,s) = 2\csc\alpha(L)\chi_{[-1,1]}(s).
$$
The natural generalization to $\nu$ would be
$$
R\nu(L) = -\sum_{x\in\partial\Omega\cap L}
\partial_{\bf n}f(x)\csc\alpha(x),
$$
where, for $x\in\partial\Omega\cap L$, we define $\alpha(x)\in[0,\pi/2]$ to be the angle of intersection between $L$ and the tangent line to $\partial\Omega$ at $x$.