Volt (V)

Divide an energy by a charge and a volt is what is left:

That comes straight from the sheet's $\Delta U_E = q\Delta V$. Move $2.0$ C through a potential difference of $9.0$ V and the electric potential energy changes by $18$ J. Move $1.0$ C through the same $9.0$ V and it changes by $9.0$ J. The potential difference did not move; the energy did.

Which is why a 9 V battery does not hold 9 J. It maintains nine joules per coulomb across its terminals, and the energy it delivers depends entirely on how much charge goes through. Treating a voltage as a quantity of energy survives the arithmetic unnoticed, because both are scalars that add, so nothing in the working looks wrong.

A volt per metre is a newton per coulomb. The sheet prints $\lvert \vec{E} \rvert = \lvert \Delta V / \Delta r \rvert$ alongside $\vec{E} = \vec{F}_E / q$, so field strength ends up with two unit expressions, and they agree: $(\text{J/C})/\text{m} = (\text{N} \cdot \text{m/C})/\text{m} = \text{N/C}$. Either is acceptable in an answer.

Volts and voltage. Voltage is loose shorthand. What the equations carry is a potential difference between two points. A bare $V$ at a point only means something once a zero is chosen, and both electricity exam-convention boxes fix that zero at an infinite distance from an isolated point charge.

Volt is printed on the AP Physics 2 and AP Physics C: Electricity and Magnetism unit-symbol tables.