Why do we run in diagonals when proving that $\mathbb{Q}$ is countable?

Whatever method you choose of indexing the rationals, it has to satisfy the following basic property: for every rational number $x/y,$ there must be some positive integer $n$ such that $x/y$ is indexed by $n$ (slightly formally: there must exist $n\in\mathbb{N}$ such that $n\mapsto x/y$).

Let's attempt what you are describing. You are saying that we should start with $1/1,$ then go to $1/2,$ then $1/3,$ and so on; "and then" move on to $2/1,$ $2/2,$ etc.

Here's my problem with that. The indexing procedure you are describing takes each positive integer $n$ and maps it to $1/n$: $1$ maps to $1/1,$ $2$ maps to $1/2,$ and so on. So, here's my question to you: under your procedure, which positive number $n$ indexes $2/1$?

Note, as I said in the first paragraph, that in order for your procedure to be a proper procedure, there must be some value of $n$ so that $2/1$ is indexed by $n.$ You claim that your procedure is a proper indexing procedure; now you have to tell me what the value of $n$ is.

I'm sure that if you meditate on this you will see what the problem is. There is no such value of $n,$ so what you are attempting to do simply will not work; your procedure does not get through all the rationals, it only ever goes through the first row of the table. This is what people mean in the comments when they say that you "run out" of integers.

Simply because you cannot "finish" the elements $\dfrac1x$, and you would never index $\dfrac2x$.