Suppose $x'_i = h_i(x_1, \ldots, x_n)$ for $i = 1,...,n$. To apply the inverse function theorem, let $f: R^{n} \rightarrow R^n $ with $f^i(x_1,...,x_n) = h_i(x_1,...,x_n)$. Assume that the determinant of f is nonzero for all $x_i$, then there will exist a function g such that $g_i(x'_1,...,x'_n) = x_i$ for $x'$ close to $(x'_1,...,x'_n)$.
But this only provides a localized inverse, right, there's no way for it to give the inverse everywhere? My classical mechanics textbook (corben) talks about this but it does not mention that the inverse is local and makes it sound like there will be a inverse that you can write down on paper. I don't see how the inverse function theorem is useful in coordinate transforms as the existence of local inverses seems pretty useless.
There's definitely no guarantee of a global inverse, as we can already see in one dimension: functions like $f(x) = x^2$ have nonvanishing derivative away from $x = 0$, and so there exist local inverses about these points, but there's no global inverse, since $x^2$ isn't one-to-one.
In any case, the IFT is of enormous theoretical importance, but less so on the practical side, at least as far as I know. It guarantees the existence of a local inverse, but says nothing about writing one down in any sort of nice way.
EDIT: It seems I have overlooked the condition that the derivative should be everywhere nonvanishing. As Jonas Meyer points out in the comments, this is actually impossible in one dimension. In two dimensions, the function $f(x,y) = (e^x \cos(y), e^x \sin(y))$ has everywhere nonsingular Jacobian, but fails to have a global inverse, as it is not injective.