Since $g(x)$ is periodic, the integral formula for its Fourier coefficients is only correct for integer index $s$. For example, we can write down the same integral, but exponentials with random coefficient in the exponent aren't orthogonal on $[0,1]$, and so on. Thus, differentiation in that parameter doesn't really have the sense we'd might have thought. The coefficient(s) $a(s)$ for $g$ are truly functions of $s\in\mathbb Z$, so differentiation has no natural sense. It is true that the $n$th Fourier coefficient of the $k$th derivative of $g$ is $(2\pi in)^k$ times the $n$th Fourier coefficient of $g$, but that's a different question.
Edit: in response to further comment... Yes, the Fourier coefficient $a(s)$ of $f$ is defined on the whole real line, and the identity about differentiating it with respect to $s$ is correct. However, to "wind up" (or whatever verb one wants) that identity does not give the integral of $g$ against $x^n$, but, instead $\int_0^1 \sum_{\ell\in\mathbb Z} (x+\ell)^n\,f(x+\ell)\,dx$. That is, the thing inside the integral is not $x^n \sum_\ell f(x+\ell)$, because the $x^n$ has to get summed over translates, too.
Edit 2: In response to second comment... it is not the case that the $n$th derivative $a^{(n)}(s)$ of $a(s)$ is $\int_0^1 x^n\,g(x)\,dx$ (with or without constants). Yes, the corresponding identity does hold with $f$ instead of $g$, but when you try to deduce the $g$-assertion from the $f$-assertion, the winding-up fails, because $x^n$ is not periodic, unlike the exponential.
Just for context, I remember when first learning calculus, thinking that I'd found an easier way to solve high-degree polynomial equations: for example, to solve $x^2=10$, _take_derivatives_ (haha!) of both sides, supposedly getting $2x=0$, so $x=0$??? No. Sure, one can take the derivative of an expression, but that does not guarantee harmony with the context, even if it sorta makes sense. The problem with the pollynomials was that $x$ is just a name for an unknown constant, so it makes no sense to take a derivative with respect to it. There's a similar problem, in part, in this Fourier business.