In the proof that weak OWFs imply strong OWFs, why must the position of $y$ be random?
As you write, the proof is via a reduction that transforms any adversary $B$ that inverts $F$ with non-negligible probability $\varepsilon(n)$ into an adversary $A$ that inverts $f$ with probability at least $1-1/p(n) \approx 1$.

Assume for simplicity that $f$ is an injective function. Consider a potential adversary $B$ that internally works as follows: given $Y=(y_1, \ldots, y_t)$, it checks whether $y_1$ belongs to a certain “special” $\varepsilon(n)$-fraction of the image $f(\{0,1\}^n)$. If so, it computes and outputs the preimage $X = F^{-1}(Y)$. Otherwise, it outputs nothing, i.e, it fails to invert this $Y$. Clearly, this $B$ satisfies the above hypothesis, because $y_1$ has an $\varepsilon(n)$ probability of being in the “special” set. (How $B$ checks $y_1$ and inverts $F$ is not important, because the reduction treats $B$ as a “block box.” So we can think of $B$ as using unbounded computation to perform these steps.)

Now, consider a reduction $A$ that works as you propose, by always “plugging in” its external challenge $y=f(x)$ in the first position, letting $y_1=y$, and choosing the rest of $Y$ itself. Clearly, $B$ outputs the preimage $F^{-1}(Y)$ if and only if $y$ belongs to the “special” set, which occurs with probability $\varepsilon(n)$. Therefore, $A$ succeeds in outputting the preimage $f^{-1}(y)$ with the same probability $\varepsilon(n)$. But this is not enough: we need $A$ to succeed with much larger probability $1-1/p(n)$.

If $A$ repeats its procedure many times, always letting $y_1=y$ but changing the other $y_i$, the probability of success would not improve: whether $B$ inverts $Y$ or not depends only on the value of $y_1=y$, and this does not change. (Remember, $A$ gets only one challenge value $y$ and needs to invert on it with high probability.)

The correct reduction and proof circumvents this problem by showing that there must exist some position $i$ for which there is a large $1-1/p(n)$ fraction of $y_i$ such that, if we choose the other $y_j = f(x_j)$ at random as described, then $B$ inverts on that $Y$ with non-negligible probability (over the choice of the $y_j$ alone). By guessing such $i$ at random (or enumerating all of them) and repeating the core procedure many times, we can ensure a high probability that $B$ actually inverts on one of the $Y$ we provide it, thereby telling $A$ a preimage of $y$. (The actual proof does not depend on $f$ being injective.)