calculus - Why does substitution work in antiderivatives?

I'm not entirely sure what I want to ask here, so please bear with me!

I think the explanation they give us in school for how finding the antiderivative by substitution works is: $$\int f(g(t))g'(t)dt=(Fg)(t)=F(g(t))=F(u)=\int f(u)du$$

But I never really understood the equality $F(u)=\int f(u)du$. Why can we behave as if the identity $u=g(t)$ doesn't exist, compute the integral $\int f(u)du$, and then 'substitute' $u$ in the result yielding the correct answer? In a sense, it feels like the variable $u$ switches from being 'meaningful' in $F(u)$ (where it stands for $g(t)$) and being insignificant in $\int f(u)du$, since as I perceive it the "variable name" here is meaningless and we could say $\int f(u)du = \int f(k)dk = \int f(x)dx = ...$ all the same. Another way to ask the same thing: why $(Fg)(t)=\int f(u)du$ and not, say, $(Fg)(t)=\int f(x)dx$? Where am I getting confused?

Maybe someone can explain this better than my high school teacher? Is there a 'formal' explanation for this? Thank you a lot!