Here's the setup: $k$ is a field extension of degree $d$ over $\mathbb{Q}$. So there are $d$ field embeddings $\sigma_1,\dotsc,\sigma_d:k \to \mathbb{C}$. Let $\{\alpha_1,\dotsc,\alpha_d\}$ be a collection of elements of $k$. The text I'm reading is (I think) trying to motivate the definition of the discriminant by posing the question of when $\{\alpha_i\}$ is a basis for $k$ as a vector space over $\mathbb{Q}$.
Of course, linear independence means considering the linear equation $$x_1\alpha_1+\dotsb+x_d\alpha_d=0$$ with coefficients $x_i \in \mathbb{Q}$. This is an equation in the abstract field $k$ so embed it in $\mathbb{C}$, I guess to make it more concrete? $$x_1\sigma_i(\alpha_1)+\dotsb+x_d\sigma_i(\alpha_d)=0$$ At this point, the text says
Thus one readily deduces that the set $\{\alpha_1,\dotsb,\alpha_d\}$ is a basis for $k$ if and only if $\det[\sigma_i(\alpha_j)] \not= 0$.
I do not readily deduce this at all to be frank. Seems to me that writing out a matrix implies a choice of basis already. What basis are they referring to? Are the $\sigma_i$ somehow acting as some sort of component functions, giving coordinates of the $\alpha_j$?
After this I'm willing to accept that it's a good idea to define the discriminant $\operatorname{disc}(\{\alpha_1,\dotsc,\alpha_d\})=\det[\sigma_i(\alpha_j)]^2$.
A homogeneous system of $n$ linear equations wth $n$ unknowns has a nontrivial solution iff the determinant of the system is $0$. That is the statement used in the OP.