Questions about voltage drop at the transformer secondary and primary due to loading
For an ideal transformer the relation 
$$ P_{in} = P_{out} \Rightarrow  P_{prim} = P_{sec} 
\Rightarrow V_{prim} \cdot I_{prim} =  V_{sec} \cdot I_{sec} $$ 
and hence
$$ \frac{V_{prim}}{V_{sec}} =  \frac{I_{sec}}{I_{prim}} $$
is always valid.

The power drawn by the secondary side influences the power that primary side (has to) provide(s).
The voltage of the primary side influences the voltage of the secondary side.
So the "primary's action is dependent from secondary".

The current drawn on the secondary side is 'provided' by the primary side. This current on the primary side causes a voltage drop in R1, reducing the voltage V1 at the primary side of the transformer. The secondary voltage changes proportionally (to the transformer ratio) with the reduced voltage V1.

You can refer the impedance at the secondary side of the transformer to the primary side of the transformer.
For the load applies
$$ R_2 = \frac{V_{2}}{I_{2}}  $$
If the load would be on the primary side, it would be
$$ R_1 = \frac{V_{1}}{I_{1}}  $$.

Using $$ \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}} $$ and $$ \frac{I_{1}}{I_{2}} = \frac{N_{2}}{N_{1}} $$
you can make define the equivilant load at the primary side being

$$ R_1 = \frac{V_{1}}{I_{1}} = \frac{ V_{2} \cdot \frac{N_{1}}{N_{2}} }{ I_{2} \cdot \frac{N_{2}}{N_{1}} } = \frac{ V_{2}}{I_{2}} \cdot \bigg( \frac{N_{1}}{N_{2}} \bigg)^2 = R2 \cdot \bigg( \frac{N_{1}}{N_{2}} \bigg)^2 $$



From the equivalent circuit it is easier to see how the load current causes a voltage drop in R1 which lowers the voltage V1.