1: Formal definitions
To start with this question, one should define a probability space: A tuple of three items usually denoted $(\Omega,\mathcal{E},\Bbb{P})$ [or something of this nature].
$\Omega$ is the sample space - the set of all possible outcomes (not to be confused with events!) of our procedure, experiment, whatever. For instance, consider flipping a coin once: In this case, $\Omega=\{\text{H},\text{T}\}$. A random variable $X$ is the "result" of this experiment. You could define $X$ in this case as
$$X=\begin{cases}
1 & \text{If coin lands heads}\\
0 & \text{If coin lands tails}
\end{cases}$$
Formally, one can define a measurement $M$ as a bijective map $M:\Omega\to\mathcal{X}$ that maps an outcome of our experiment to a value of the random variable. Here $\mathcal{X}$ is the set of all possible values of $X$. In this coin case, the "measurement" could be writing down a $0$ or $1$ in your notebook if you see a tails or heads accordingly. Bijective means one-to-one: No two outcomes can have the same measurement, and no two measurements could have come from the same outcome.
$\mathcal{E}$ is the event space, which is the set of all subsets (or powerset) of the sample space $\Omega$. In set notation, $\mathcal{E}=\mathcal{P}(\Omega).$ In the coin case mentioned above, $\mathcal{E}=\{\varnothing,\{\text{H}\},\{\text{T}\},\{\text{H},\text{T}\}\}$.
$\mathbb{P}$ is a probability function or probability measure, which is a map or function that maps an event in the event space to a probability. Formally, $\mathbb{P}:\mathcal{E}\to[0,1].$ $\Bbb{P}$ always satisfies three conditions:
1: $\Bbb{P}(e)\in[0,1]~\forall e\in\mathcal{E}$
2: $\Bbb{P}(\varnothing)=0.$
3: $\Bbb{P}(\Omega)=1.$
In words, 1: Every event has a probability. 2: Our experiment must have a result, or, the probability of nothing happening is $0$. 3: Something will happen, or, the probability of getting any result is $1$.
2: Distributions
A probability distribution is a map or function $p$ that assigns a number (positive or zero), not necessarily between $0$ and $1$, to every possible value of $X$. Formally, $p:\mathcal{X}\to\Bbb{R}_{\geq 0}$. In the discrete case, it is quite closely related to the probability measure mentioned before. Let $x\in\mathcal{X}$ be the result of a measurement of some possible outcome, say $x=M(\omega)$ for some $\omega\in\Omega$. It actually turns out that in the discrete case,
$$p(x)=\Bbb{P}(\omega).$$
So one might ask: what is the difference between these two closely related things? Well, note that in the continuous case, the above equality does not hold. Since $\Omega$ is uncountably infinite, the probability of any single outcome, or indeed any countable subset of outcomes, is zero. That is,
$$\mathbb{P}(\omega)=0$$
regardless of the value of $p(x)$.
In the discrete case, $p$ must satisfy the condition
$$\sum_{x\in\mathcal{X}}p(x)=1$$
And in the continuous case
$$\int_{\mathcal{X}}p(x)\mathrm{d}x=1$$
How can we interpret the value of $p(x)$? In the discrete case this is rather simple: $p(x)$ is the probability of measuring a value $x$ from out experiment. That is,
$$p(x)=\mathbb{P}(X=x).$$
But in the continuous case, one must be more careful with how we interpret things. Consider two possible measurements $x_1$ and $x_2$. If $p(x_1)>p(x_2)$, then $\exists\delta>0$ such that $\forall\epsilon<\delta$ (with $\epsilon>0$),
$$\Bbb{P}(X\in[x_1-\epsilon,x_1+\epsilon])>\Bbb{P}(X\in[x_2-\epsilon,x_2+\epsilon])$$
In simple terms, we are more likely to measure a value close to $x_1$ than close to $x_2$.
I would recommend watching 3Blue1Brown's video about probability density functions.