The answer to your question becomes quite easy if you are able to build the correct mathematical framework. Note that I try to build an answer adapted to the OP background, whence it will not be strictly rigorous.
First of all, let me try to explain the definition of the delta "function". In mathematics, the Dirac delta is an example of what we call distributions (or generalized functions), roughly speaking mappings (functionals) that assign to each smooth function a real number; in other words $T$ is a distribution if
$$ T : \{ \text{smooth functions vanishing at infinity} \} \rightarrow \mathbb{R}, $$
it is linear and has a continuity property I won't write explicitly. The vanishing at infinity condition should also be understood in a suitable sense, but let me go on. For a fixed $\mathbf{r}_0 \in \mathbb{R}^3$, the Dirac delta $\delta_{\mathbf{r}_0}$ acts on smooth functions $f : \mathbb R^3 \rightarrow \mathbb R$ as
$$\delta_{\mathbf{r}_0}(f) = \langle \delta_{\mathbf{r}_0}, f \rangle = f(\mathbf{r}_0) \in \mathbb{R}. \qquad (*) $$
Note that the smoothness of $f$ ensures that the pointwise evaluation makes sense. This reminds the last identity you wrote in the question, with the bracket notation (which is common in the theory of distributions) instead of the integral since the Delta is not a function in the common sense, thus we can't integrate it in a classical sense.
Other examples are distributions defined by means of an integration, for instance
$$ f \mapsto \int_{\mathbb{R}^3} f g \quad \text{or } \ \ \ f \mapsto \int_{\mathbb{R}^3} (\partial_{xy}f)g, $$
where say $g$ is another fixed integrable function (note that here the condition at infinity can be crucial).
So what's interesting about this concept? The fact that we can do some calculus with distributions, in particular there is a notion of derivative for distributions which yields other distributions! The idea is rather simple: since distributions act on smooth functions that can be differentiated as many times as we like, we define the derivative of a given distribution dumping the derivation on the smooth function. If $T$ is any distribution, $\partial_x T$ (same for other derivatives) is the ditribution defined via:
$$(\partial_x T) (f) := - T (\partial_x f); \qquad (**) $$
notice that the evaluation on the right hand side has a meaning since $T$ is a distribution and $\partial_x f$ is smooth because $f$ is.
We already have the tools to compute the derivative of the Dirac delta, indeed by $(*)$ and $(**)$:
$$(\partial_x \delta_{\mathbf{r}_0})(f) = - \delta_{\mathbf{r}_0}(\partial_x f) = -(\partial_x f)(\mathbf{r}_0).$$
The notion of gradient of a distribution (and hence of a Dirac delta) is quite similar to $(**)$, although not completely intuitive since such gradient is not a "vector", it rather acts on vectors (i.e., colletions of smooth functions!):
$$\nabla T : \{ \text{smooth functions vanishing at infinity} \}^3 \rightarrow \mathbb{R}, \qquad \nabla T (f_1, f_2, f_3) := - ( T(\partial_x f_1) + T(\partial_y f_2) +T(\partial_z f_3) ).$$
Now you can try to put the pieces together and compute the (distributional) gradient of a Dirac delta. The notions of curl and divergence can be defined in a similar way, and for the Delta will result in distributions that act as combinations of pointwise evaluations of the derivatives of the "test function/vector" $f$.
