The distribution is simply the assignment of probabilities to sets of possible values of the random variable. If I tell you how probable it is that a certain random variable is between $3$ and $5$, and also how probably it is that it's in every other possible set, then I've told you the distribution. Since I can't do this for every set individually, since there are infinitely many sets, perhaps a more down-to-earth way to say this is this: Suppose $X$ and $Y$ are random variables. If it is true of every set that the probability that $X$ is in that set is the same as the probability that $Y$ is in that same set, then $X$ and $Y$ have the same distribution.
A probability density function is a way of characterizing some distributions. For example, consider the function
$$
f(x) = \begin{cases} 0 & \text{if }x<0, \\ e^{-x} & \text{if }x\ge 0. \end{cases}
$$
To say that this is the probability density function of a random variable $X$ is to say that for every measurable set $A$ of real numbers,
$$
\Pr(X\in A) = \int_A f(x)\,dx.
$$
The probability assigned to each set $A$ is given by the integral above. A more concrete example:
$$
\Pr(3<X<5) = \int_3^5 e^{-x}\,dx\text{ and }\Pr(X\ge 2) = \int_2^\infty e^{-x}\,dx.
$$
Not every probability distribution has a density. Say we let $X$ be the number of aces when a die is thrown four times. Then $X\in\{0,1,2,3,4\}$. The probability distribution assigns a positive number to every set that intersects that last set. For example the set $\{x : x\ge 3.2\}$ intersects $\{0,1,2,3,4\}$ and thus the probability distribution of $X$ assigns a positive number to that set. But there is no function $f$ such that for every set $A$ we have $\int_A f(x)\,dx$ equal to the probability that $X\in A$.
PS prompted by comments below: To put it in a different kind of language: Say $m$ is a measure (not necessarily assigning finite measure to the whole space) on the set of all measurable subsets of a space $S$. A probability density with respect to the measure $m$ is a measurable function $f:S\to[0,\infty)$ such that the function
$$
A\mapsto \int_A f\,dm
$$
is a probability measure on the set of measurable subsets of $S$.
A probability distribution on $S$ is simply a probability measure on the set of all measurable subsets of $S$. But not quite "simply": The probability distribution of a random variable $X:\Omega\to S$ is the probability measure on measurable subsets of $S$ that assigns measure $P(\{\omega\in\Omega : X(\omega)\in A\})$ to each measurable subset $A$ of $S$.
PPS: When $f\ge0$ is a measurable function on Borel or Lebesgue-measurable subsets of $\mathbb R$, one sometimes refers to the "measure" $f(x)\,dx$, meaning the measure
$$
A\mapsto \int_A f(x)\,dx.
$$
If in addition $\displaystyle\int_{\mathbb R} f(x)\,dx=1$, so that $f$ is a probability density, then one may similarly refer to the "probability distribution" $f(x)\,dx$.
(Of course, not all probability distributions on Borel subsets of the real line are of this form.)
Best Answer
Almost. The correct density w.r.t. $\mathcal{L}_0$ would be $$h:x \mapsto (1-\alpha)g(x) + (\alpha - (1-\alpha)g(0)) \delta_0(x)$$ We can check this by noting for any Borel set $A\subseteq \mathbb{R}$, that $$\int_A h(x) \mathcal{L}(dx) =(1-\alpha)G(A)$$ and $$\int_A h(x) \: \bar{\delta}_0(dx) = \alpha \bar{\delta}_0(A),$$ which would give that $$\int_A h(x) \: \mathcal{L}_0 (dx) = (1-\alpha)G(A) + \alpha \bar{\delta}_0(A)=\mathbb{P}_X(A).$$