Space complexity of Horner's method
Storing the output requires roughly $n|x|$ bits (where $|x|$ is the size of $x$ in bits); this is since the numbers involved have magnitude $\Theta(x^n)$. This is all that Horner's method requires, since it only keeps track of one value, the current "partial sum".

In contrast, the naive method keeps track of two values: the current partial sum, and also the current power of $x$. Each of these requires (towards the end of the process) roughly $n|x|$ bits to store, and both of them require $2n|x|$ bits in total.