and not necessarily being
$$I_2=\frac{V_1I_1}{V_2}$$
In fact, assuming an ideal transformer, this is necessary. There is no contradiction for a resistive load.
Assuming an ideal transformer of turns ratio $n = \frac{V_p}{V_s} = \frac{I_s}{I_p}$ and a resistive load with resistance $R_L$, there is only one independent variable.
If the independent variable is the primary voltage then:
$$I_p = \frac{V_p}{n^2 R_L}$$
$$I_s = nI_p = \frac{V_p}{nR_L}$$
$$V_s = I_sR_L = \frac{V_p}{n} $$
If the independent variable is the primary current then:
$$V_p = I_pn^2 R_L$$
$$V_s = \frac{V_p}{n} = I_pnR_L$$
$$I_s = \frac{V_s}{R_L} = nI_p$$
Normally, do we apply an alternating voltage on the primary coil or an
alternating current?
In the case of AC power transformers, the AC mains are very good approximations to a voltage source with very low output impedance.
In the case of audio output transformers used in vacuum tube amplifiers, the primary typically sees the relatively high output impedance looking into the anode of the output tube(s).
So, the answer is it depends on the application.
But, as stated above, for a resistive load the voltage across and current through the primary are proportional and, as The Photon has noted, "you can't have one without the other".