Suppose you have a complex Radon measure $\mu$, treated as a distribution. Then does every such Radon measure admit a decomposition of the form
$\mu = \sum_{n=1}^\infty c_n \delta(x-\tau_n) + \hat f$
where the $c_n$ are complex numbers, the $\tau_n$ are real numbers, and $\hat f$ is some ordinary function, treated as a distribution?
The Cantor distribution shows that the singular part of a Radon-Nikodym decomposition need not be a sum of point masses.
Let $F: [0,1] \to [0,1]$ be the Cantor function. Since $F$ is monotone increasing, we can construct a measure $\mu$ on the Borel sets of $[0,1]$ by defining $\mu(\varnothing) = 0$, $\mu(\{0\}) = F(0) = 0$, and $\mu((a,b]) = F(b) - F(a)$ for $(a,b] \subseteq [0,1]$. By a standard argument $\mu$ extends to a regular Borel measure on $[0,1]$. This construction can be found in Royden's Real Analysis text.
One can show that $\mu$ has the following properties:
$\mu([0,1]) = 1$, so $\mu$ is nontrivial.
The Radon-Nikodym derivative of $\mu$ with respect to Lebesgue measure is the usual derivative of $F$, which is zero almost everywhere.
$\mu(\{p\}) = 0$ for every $p \in [0,1]$.
In your decomposition, property 2 implies $\hat{f} = 0$ and property 3 implies $c_n$ is zero for any choice of $\tau_n$, so $\mu$ would have to be the zero measure. Since $\mu$ is not the zero measure, this decomposition does not hold in general.