I'm trying to compute the joint CDF for the following PDF,
$$ f(y_1, y_2) = \begin{cases} e^{-y_1}, &\quad 0 \leq y_2 \leq y_1 < +\infty \\ 0, &\quad \text{otherwise} \end{cases} $$
with the following the $X = Y_1 - Y_2$.
I compute, \begin{align} P(X \leq x) &= \int \int_{\{y_1, y_2: X \leq x\}} f(y_1, y_2) dy_1 dy_2 \\ &= \int \int_{\{y_1, y_2: Y_1 \leq x + y_2\}} f(y_1, y_2) dy_1 dy_2 \\ &= \int_{y_1 = y_2}^{y_1 = x + y_2} \int_{y_2 = 0}^{y_2 = y_2} e^{-y_1} dy_1 dy_2 \end{align}
However it seems like the second integral should go until $y_2 = +\infty$ but I don't understand why as the definition of the CDF is, $$ P(Z \leq z) := \int_{0}^{z} f(z)dz $$ for any random variable $Z$.
You have the integral trees backwards.
Using your notation, you should have. $$\begin{align}\mathsf P(X\leq x) &= \int_{y_2=0}^{y_2\to\infty}\left.\int_{y_1=y_2}^{y_1=y_2+x}\mathrm e^{-y_1}\,\mathrm d y_1\right.\,\mathrm d y_2\end{align}$$
We are integrating over all supported values for $y_2$, and inside that we are integrating over the conditionally supported values for $y_1$ where $y_1-y_2\leq x$.$$0\leq y_2~\leq y_1\leq y_2+x $$
Note: $y_1$ cannot be mentioned inside the domain for the outer integral, because it is bound inside the scope of the inner integral.
$$\begin{align}\mathsf P(X\leq x) &=\int_{s\leq x} f_{X}(s)\,\mathrm d s\\&= \int_{s\leq x} f_{Y_1-Y_2}(s)\,\mathrm d s\\&=\iint_{s\leq x} f_{Y_1-Y_2,Y_2}(s,y_2)\,\mathrm d s\,\mathrm d y_2\\&=\iint_{s\leq x} f_{Y_1,Y_2}(s+y_2,y_2)\,\mathrm d s\,\mathrm d y_2\\&=\iint_{y_1\leq y_2+x} f_{Y_1,Y_2}(y_1,y_2)\,\mathrm d y_1\,\mathrm d y_2\\&=\iint_{0\leq y_2\leq y_1\leq y_2+x} \mathrm e^{-y_1}\,\mathrm d y_1\,\mathrm d y_2\\&=\int_{0\leq y_2}\int_{y_2\leq y_1\leq y_2+x}\mathrm e^{-y_1}\,\mathrm d y_1\,\mathrm d y_2\\&=\int_{y_2=0}^{y_2\to\infty}\int_{y_1=y_2}^{y_1=y_2+x}\mathrm e^{-y_1}\,\mathrm d y_1\,\mathrm d y_2\end{align}$$
In short: The domain of integration must cover the region where $y_1-y_2\leq x$ and $0\leq y_2\leq y_1<\infty$ .