Why does the bounds of this integral not consider both equalities?

When you write a definite integral such as

$$ \int_a^b f(x,y)\, \mathrm dy, $$

The variable $y$ wherever it appears in that integral comes from the $y$ in $\mathrm dy.$ All of the instances of the variable named $y$ inside the integral are invisible and inaccessible to any expressions outside the integral. In fact, technically the choice of the name $y$ is irrelevant and any variable name that is not already in use in the integral could have been chosen:

$$ \int_a^b f(x,y)\, \mathrm dy = \int_a^b f(x,z)\, \mathrm dz = \int_a^b f(x,v)\, \mathrm dv. $$

The fact that $y$ was chosen in the written solution is merely a helpful hint to you that this $y$ is somehow related to the $y$ in the definition of the distribution.

So the idea that we could somehow use $y$ outside the inner integral, for example to set bounds of the variable $x$ in the outer integral, simply will not work.

Fortunately, when we set the boundary values of the inner integral as shown in the solution,

$$ \int_0^x f(x,y)\, \mathrm dy, $$

this enforces the condition that $y < x$ by requiring $y$ to be integrated only over values between $0$ and $x.$ This is enough. The condition, once enforced, stays enforced.

Intuitively:

$$\begin{align}\iint_{\text{all supported values}}\dfrac 1x \,\mathrm d\langle x,y\rangle &= \int_{\text{all values $x$ may take}}\left[\int_{\text{all values $y$ may take for a particular $x$}}\dfrac 1x\,\mathrm d y\right]\,\mathrm d x\\[2ex]\iint_{0\leq y\leq x\leq 1}\dfrac 1x \,\mathrm d\langle x,y\rangle &= \int_{0\leq x\leq 1}\left[\int_{0\leq y\leq x}\dfrac 1x\,\mathrm d y\right]\,\mathrm d x\\[1ex]&= \int_0^1\left[\int_0^x\dfrac 1x\,\mathrm d y\right]\,\mathrm d x\end{align}$$

Because the domain is $\{\langle x,y\rangle: 0\leq y\leq x\leq 1\}=\{\langle x,y\rangle: 0\leq x\leq 1\text{ and }0\leq y\leq x\}$

Consider a count of the integer sequence $\{(1,1),(2,1),(2,2),(3,1),(3,2),(3,3)\}$ which is $\{\langle x,y\rangle\in\Bbb N^2:1\leq y\leq x\leq 3\}$. This is clearly 6, and taking the series agrees:

$$\begin{align}\sum_{\small\langle x,y\rangle\in\Bbb N^2:1\leq y\leq x\leq 3}1&=\sum_{x=1}^3\sum_{y=1}^x1\\&=\sum_{x=1}^3x\\&=6\end{align}$$

But if we do not "sum out" the inner variable and bound the outer series as $\{x\in\Bbb N:y\leq x\leq 3\}$:

$$\begin{align}\sum_{x=y}^3\sum_{y=1}^x 1&=\sum_{x=y}^3x\\&=y+\ldots+3\end{align}$$

Which is clearly incorrect, as we have left $y$ as a free variable.

The same principle applies.