Two coils on one iron core, and a changing current in the first induces a voltage in the second. The proportion between them is set by nothing more exotic than how many times each wire went round.

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ — V-s over V-p equals N-s over N-p. Fix the subscripts once and the rest is bookkeeping: p is the primary, the winding you energise, and s is the secondary, the winding you take power from. $V_p$ and $V_s$ are those windings' voltages in volts; $N_p$ and $N_s$ are their turn counts, plain numbers with no unit. Note that this $N$ counts turns of wire — it is not the $N_s$ of the first lesson, which was a speed in rpm. Same letter, different machine, and the sentence around it tells you which.

Halve the turns and you halve the volts. What the ideal transformer will NOT do is halve the power, so the current does the opposite: step the voltage down and the current steps up, by the same ratio. A 10:1 step-down with 10 A in the primary delivers about 100 A out of the secondary, which is exactly why the secondary bar is the fat one and why reading the ratio backwards is the marquee trap in this lesson.

The other half of the nameplate is the rating. A transformer is rated in kVA, not kW, because its losses depend on current and voltage and not on what the load's power factor happens to be. From it, $I_{FL} = \dfrac{S}{\sqrt{3}\, V}$ for a three-phase unit — $S$ the rating in volt-amperes, $V$ the line-to-line voltage of the winding you are asking about, $I_{FL}$ that winding's rated current. Watch the kilo: the S box wants VA, and a nameplate speaks kVA.