This is from chapter 3.6 of the textbook by John Rice on probability and statistics:
The continuous case is very similar. Supposing that $X$ and $Y$ are continuous random variables, we first find the cdf of $Z$ and then differentiate to find the density. Since $Z \leq z$ whenever the point $(X, Y)$ is in the shaded region $R_z$ shown in Figure $3.17$, we have \begin{align} F_Z (z) &=\iint\limits_{R_z} f(x,y) \ dx \ dy \\ &= \int_{-\infty}^\infty \int_{-\infty}^{z-x} f(x,y) \ dy \ dx. \end{align} In the inner integral, we make the change of variables $y=v-x$ to obtain \begin{align} F_Z (z) &= \int_{-\infty}^\infty \int_{-\infty}^z f(x,v-x) \ dv \ dx \\ &=\int_{-\infty}^z \int_{-\infty}^\infty f(x,v-x) \ dx \ dv. \end{align} Differentiating, we have, if $\int_{-\infty}^\infty f(x,z-x) \ dx$ is continuous at $z$, $$f_Z (z) =\int_{-\infty}^\infty f(x,z-x) \ dx.$$
I don't understand how they were applied the change of variables in the double integral. Why does the upper bound of the inner integral change from $z-x$ to $z$?
Because when $v=z$ you have $y=v-x = z-x$ as like in the first integral.