Ohm's law interpretation correction

Generally, we see current as a response to voltage- there is movement of electrons due to a potential difference between the ends of a conductor. (Voltage difference means a potential drop across the ends of a conductor. Electrons flow from a region of higher potential to a region of lower potential.) As a physicist, $ I∝V $ feels more intuitive.
From a mathematical point of view, both are correct, as you said. But if I were an electrician, looking to put wires in a building, I would first check the amount of electric current needed by the equipment, and then decide the voltage. So here, you could say that the potential difference (voltage drop) would be directly proportional to the current (practically) here. (I do not know if electricians actually work this way. This was meant as an example.)
As a cause-effect relationship, voltage drop is considered the cause and the current as the effect, so I would say, $I \propto V$ is a better way of thinking about it.

They mean mostly the same. One may see voltage$(V)$ as a measure of electric field in the wire. Electric field is what applied force on the charge carriers (in case of conductors, it's electrons). And since the flow of charge carriers per unit time is current$(I)$, we say that:

1. For greater current, there should be greater voltage. $$V\propto I$$ As you mentioned, the proportionality constant here would be $R$. And, this is known as resistance. And, this is a property of the material to resist the flow of current. So, if the resistance is high, greater electric field(in turn voltage) is needed for the same current to flow, hence the relation: $$V=RI,$$ which clearly demonstrates this.

OR, we can say:

2. Greater the voltage greater the current i.e. $$I\propto V$$ Here, the proportionality constant here would be $1/R(=G)$. And, this is known as conductance. Current, other than depending than depending on electric field(voltage), also depends of the ease of flow(opposite of resistance, hence reciprocal) of charge carriers through the material. So, greater the ease greater the current for the same voltage i.e. $$I=\frac{1}{R}V=G V$$.

So, the interpretation is clearly simple to the point, both relations follows just from $V=IR$ as:

$$ V=IR\implies I=\frac{V}{R} =GV $$

Deviation from Ohm's law

While Ohm's law is valid for a large range of materials, it fails for some i.e. the linear relation between current and voltage is not followed and higher order terms like $I^2$ or $V^n$ etc. come around along with other constants. A notable and famous example is doped semiconductors (such as, Gallium arsenide).

The order does not matter except when plotting experimental data.
The convention is to plot the dependent variable along the y-axis and the independent variable along the x-axis resulting in, $$\text{dependent variable }\propto \text{independent variable}$$

Thus it all depends if you are fixing the value of voltage (or current) and measuring the resultant current (or voltage).

The current-voltage characteristics of a circuit element (not only resistors) are usually plotted with the current as the dependent variable and the voltage as the independent variable.

The vector form of Ohm's law is usually written as $\vec J = \sigma \, \vec E$, where $\vec J$ is the current density, a vector representing the amount of current flowing per unit area, $\sigma$ is the electrical conductivity of the material, and is the reciprocal of the resistivity and $\vec E$ is the electric field, a vector representing the force per unit charge.
The scalar equivalent of this relationship is $I= \dfrac 1R\cdot V$, ie $I\propto V$.