Why do we use square in measuring a qubit with probability? A superposition is as follow:
$$ \vert\psi\vert = \alpha\vert 0\rangle + \beta\vert 1\rangle. $$

When we measure a qubit we get either the result 0, with probability $\vert \alpha\vert^{2},$ or the result 1, with probability $\vert \beta\vert^{2}.$
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Why does the $\alpha$ have the power $2$ ? 

Depending on how you are setting up your quantum states, it is usually taken as an axiom of the system that if a state $|\psi\rangle$ is a linear superposition of eigenstates $\{ |e_n\rangle \}$ of some observable where 
$$|\psi\rangle = \sum_n \alpha_n |e_i\rangle \ ,  \ \alpha_n \in \mathbb C \text{ and the coefficients are normalized: } \sum_n |\alpha_n|^2 = 1$$
then when we make a measurement with respect to that observable, the state is observed in state $|e_i\rangle$ with probability $|\alpha_i|^2$. Whatever else the probability is, it cannot be $\alpha_i$, as that is complex. $|\alpha_i|$ is a possible choice. But it will turn out that the mathematics and physics of the model are much more fruitful if we say the probability is $|\alpha_i|^2$.
In other words, it's a model which has made an arbitrary choice about how to interpret it and use it. And it turns out it's a very successful model.
(To be convinced of that last point, keep on using the model in your course or elsewhere!)
