I am reading "Analysis on Manifolds" by James R. Munkres.
Definition. Let $k\leq n$. Let $A$ be open in $\mathbb{R}^k$, and let $\alpha:A\to\mathbb{R}^n$ be a map of class $C^r (r\geq 1)$. The set $Y=\alpha(A)$, together with the map $\alpha$, constitute what is called parametrized-manifold, of dimension $k$. We denote this parametrized-manifold by $Y_\alpha$; and we define the ($k$-dimensional) volume of $Y_\alpha$ by the equation $$v(Y_\alpha)=\int_A V(D\alpha),$$ provided the integral exists.
Definintion. Let $A$ be open in $\mathbb{R}^k$; let $\alpha:A\to\mathbb{R}^n$ be of class $C^r$; let $Y=\alpha(A)$. Let $f$ be a real-valued continuous function defined at each point of $Y$. We define the integral of $f$ over $Y_\alpha$, with respect to volume, by the equation $$\int_{Y_\alpha} f\mathrm{d}V=\int_A (f\circ\alpha)V(D\alpha),$$ provided this integral exists.
Here we are reverting to "calculus notation" in using the meaningless symbol $\mathrm{d}V$ to denote the "integral with respect to volume." Note that in this notation, $$v(Y_\alpha)=\int_{Y_\alpha}\mathrm{d}V.$$ We show that this integral is "invariant under reparametrization."
Theorem 22.1. Let $g:A\to B$ be a diffeomorphism of open sets in $\mathbb{R}^k$. Let $\beta:B\to\mathbb{R}^n$ be a map of class $C^r$; let $Y=\beta(B)$. Let $\alpha=\beta\circ g$; then $\alpha:A\to\mathbb{R}^n$ and $Y=\alpha(A)$. If $f:Y\to\mathbb{R}$ is a continuous function, then $f$ is integrable over $Y_\beta$ if and only if it is integrable over $Y_\alpha$; in this case $$\int_{Y_\alpha} f \mathrm{d}V=\int_{Y_\beta} f \mathrm{d}V.$$ In particular, $v(Y_\alpha)=v(Y_\beta)$.
What is a reparametrization?
In Theorem 22.1, which is a reparametrization, $\alpha$ or $\beta$?
Of course, since $g^{-1}:B\to A$ is also a diffeomorphism of open sets in $\mathbb{R}^k$, it is not important which is a reparametrization.
Let $Y$ be a subset of $\mathbb{R}^n$.
Let $A\subset\mathbb{R}^k$ be an open set.
Let $B\subset\mathbb{R}^k$ be an open set.
Let $\alpha:A\to\mathbb{R}^n$ be a map of class $C^r$ such that $Y=\alpha(A)$.
Let $\beta:B\to\mathbb{R}^n$ be a map of class $C^r$ such that $Y=\beta(B)$.
We don't say such $\alpha$ is a reparametrization.
We don't say such $\beta$ is a reparametrization.
Why?
Perhaps the best way to answer this is with an example.
Everything here is smooth, i.e. $C^\infty$. The map $\alpha: (-1, 1) \to \mathbb{R}^2$ given by $\alpha(t) = (t, \sqrt{1 - t^2})$ gives a parametrization of the (open, upper, unit) semicircle $x^2 + y^2 = 1,\, y>0$.
Now, the map $g: (0, \pi) \to (-1, 1)$ defined by $g(\theta) = \cos \theta$ provides a diffeomorphism $(0, \pi) \to (-1, 1)$.
This composition $\beta = \alpha \circ g: (0, \pi) \to \mathbb{R}^2$ is a reparametrization of $\alpha$, as it describes the same curve ($1$-manifold), i.e. the same set of points in $\mathbb{R}^2$, but the particular way that each value of the parameter describes the points has changed. For instance, the point $(x, y) = \bigl(\tfrac12, \smash{\tfrac{\sqrt3}{2}}\bigr)$ is the image of $t = \tfrac12$ under $\alpha$, but it's the image of $\theta = \frac\pi3$ under $\beta$.
The point being made in the text is that whenever you change variables in an integral over a parametrized manifold, you're actually just working with a different parametrization of the same manifold. The formulas will look different, but the value of the integral will not change.
Addendum 1: Switching perspective, it's also true that $\alpha = \beta \circ g^{-1}$ gives a reparametrization of $\beta$ since we can invert the diffeomorphism $g$ that ties them together. In fact, this relation among parametrizations is not only symmetric, but also reflexive (use identity diffeo) and transitive (use composition of diffeos), so we have an equivalence relation, and the manifold itself can be identified with the equivalence class of all its parametrizations.
Addendum 2: An even more familiar place where we reparametrize all the time (although this doesn't involve manifolds) is when we want to describe a discrete family of integers, say those that are congruent to $1 \bmod{4}$. We can parametrize these as $\alpha: \mathbb{Z} \to \mathbb{Z},\; \alpha(k) = 4k + 1$. But we might want the the first (index $1$) to be $1$, so we need $\beta(m) = 4m - 3$. In this case, the bijection $g:\mathbb{Z} \to \mathbb{Z}$ is given by $g(m) = m - 1$. In this context, this is usually called reindexing, but it's morally the same thing. And if we want to sum over this set, the reindexed sum will look different but give the same value.