why does a step up transformer cause lower currents?
Ohm's law is not violated because a transformer is not a resistor. The transformer simply changes the source's view of the load impedance. So if you have a 1:2 transformer connected to a source and a load resistor, then the source side would look like one quarter of the load resistance.

Transformers don't work at DC and hence understanding what's going on requires a bit more math and physics.  Let's take a simple example circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

In this case, we have a source (V1), a load (R1) and a 1:2 step-up transformer (N = 0.5).  Some current will flow on the primary side of the transformer, \$i_p(t)\$.  The current in the primary will induce a magnetic field in the transformer core that's proportional to the current and to the number of turns in the primary:

$$B_p(t) = \mu_0 k_1 N_p i_p(t)$$

Don't worry about the \$k_n\$ constants, they're dependent on the construction of the transformer and they'll cancel out at the end, assuming the transformer is ideal.  This results in a magnetic flux on the secondary:

$$\phi_s(t) = k_2 B_p(t) = \mu_0 k_1 k_2 N_p i_p(t)$$

The oscillating magnetic flux will induce a voltage in the secondary of the transformer that's proportional to the derivative of the magnetic flux and to the number of turns in the secondary:

$$v_s(t) = N_s \frac{d}{dt} \phi_s(t) = \mu_0 k_1 k_2 N_p N_s \frac{d}{dt}i_p(t)$$

The same thing can be calculated in the other direction:

$$v_p(t) = N_p \frac{d}{dt} \phi_p(t) = \mu_0 k_1 k_2 N_p N_s \frac{d}{dt}i_s(t)$$

If we assume an ideal transformer with 100% coupling between the primary and secondary, then \$\phi_s(t) = \phi_p(t)\$.  If \$\phi_s(t) = \phi_p(t)\$, then \$\frac{d}{dt}\phi_s(t) = \frac{d}{dt}\phi_p(t)\$.  This means we can write:

$$\phi_s(t) = \phi_p(t) = \mu_0 k_1 k_2 N_p i_p(t) = \mu_0 k_1 k_2 N_s i_s(t)$$

and

$$\frac{d}{dt}\phi_s(t) = \frac{d}{dt}\phi_p(t) = \frac{v_p(t)}{N_p} = \frac{v_s(t)}{N_s}$$

from which you can write

$$i_s(t) = \frac{N_p}{N_s}i_p(t)$$

and

$$v_s(t) = \frac{N_s}{N_p}v_p(t)$$

If \$N_s = 2 N_p\$ as in the example circuit, then \$v_s(t) = 2v_p(t)\$ and \$i_s(t) = \frac{1}{2}i_p(t)\$.  This makes sense from a standpoint of conservation of energy; \$P = v_s i_s = 2 v_p \frac{1}{2} i_p = v_p i_p\$.  

Ohm's law on the load resistor indicates that \$v_s(t) = 100\ \Omega\cdot\ i_s(t)\$.  By plugging in both current and voltage relations, we get \$v_p(t) = 25\ \Omega\cdot\ i_p(t)\$.  In effect, the transformer makes the 100&nbsp;&Omega; resistor 'look like' a 25&nbsp;&Omega; resistor from the standpoint of the source.  Again, this makes sense - half the voltage and double the current is 1/4 the resistance.  In general a transformer will 'transform' the load impedance by a factor of \$N_p^2/N_s^2\$