Here we have a bunch of positively charged particles (colored black) and negatively charged particles (colored white):
Now suppose we drop in a negatively charged particle at point A. It's going to try to move left, because it's attracted by all those positive charges on the left and repelled by the negative charges on the right. (There's also a negative charge on the left, but that's more than balanced out by all the positives.)
Suppose you want to move that particle from point A to point B. Then you're going to have to push against all that electrical force, so it will take some energy to move that charge from A to B.
The voltage between points A and B is the amount of energy you'll need for that --- that is, the amount of energy it takes to move your negative charge from A to B, overcoming the electrical forces along the way.
Suppose that voltage happens to be, say, 3. One way to express that is to say that the voltage at A is 1 and the voltage at B is 4. Or you can say that the voltage at A is 6 and the voltage at B is 9. Or that the voltage at A is $-2$ and the voltage at B is $+1$. You can pick a perfectly arbitrary number to assign to point $A$, as long as you assign that number plus 3 to point $B$.
So let's go ahead and say (arbitrarily) that the voltage at $A$ is $2$ and the voltage at $B$ is $5$. Again, all we mean by this is that it takes 3 units of energy to move one unit of charge from $A$ to $B$.
Now suppose there's another point $C$, and suppose it takes 7 units of energy to move a unit of charge from $A$ to $C$. That is, the voltage from $A$ to $C$ is $7$. Then since we already decided to call the voltage $2$ at point $A$, we have to call it $9$ at point $C$.
Now: How much energy does it take to move a unit of charge from $B$ to $C$? Well, the number we assigned to $B$ --- the voltage at $B$ --- is $5$. And the voltage at $C$ is $9$. Therefore, we predict that it will take $9-5=4$ units of energy to move a unit of charge from $B$ to $C$. And it turns out, empirically, that if you make predictions this way, you're always right.
So in summary: The voltage between $A$ and $B$ is the energy needed to move a unit charge from $A$ to $B$. The voltage at $A$ is any number you care to make up --- you can call it $2$ or $-100$ or $3.14159$. Once you've made that number up, the voltage at $B$ or $C$ or $D$, minus the voltage at $A$, is the energy needed to move a unit charge from $A$ to $B$ or $C$ or $D$. And---miraculously---once you assign numbers this way, you can also use them to figure out how much energy it takes to move a unit charge from $B$ to $C$ or from $B$ to $D$ or from $D$ to $C$, just by taking differences.