What is uniformity in Boolean circuits exactly?
Here is the binary encoding of 5: $$101$$ And here is its unary encoding: $$11111$$ Hopefully the difference is apparent.
Let's see what happens if we switch to binary encoding in the context of the linked question. Given the binary encoding of $n$, we need to produce a circuit, operating on $n$ bits, in logspace, and so in polynomial time. How much time do we have? The binary encoding of $n$ has length roughly $\log n$, so we have time $O(\log^C n)$, which isn't enough to produce a circuit which reads all input bits.
In contrast, if we use unary encoding, then the length of the input is $n$, and so we have time $O(n^C)$, which is more reasonable (in addition, we are only allowed to use $O(\log n)$ space).