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 $x$.