I am looking for results on the probabilities of common roots of $f(x)$ and $f'(x)$ for $f(x) = \sum_{s \in S_f} u_s x^s$ and $u_s$ i.i.d $\mathcal{N}(0,1)$-distributed, for $x \in [0,1]$. Or, put differently, how likely is it for a root of $f(x)$ to have multiplicity larger than 1?
Thank you for any ideas or reading suggestions.
You did not let us know what $S_f$ is. Since the title of your post is "Multiple roots of Random Polynomials", it seems reasonable to assume that $S_f$ is the set of integers $n_0,\dots,n_k$ such that $0\le n_0<\dots<n_k$.
One of the following cases must occur:
Case 1: $n_0\ge2$. Then $0$ is a multiple root of $f$ with probability $1$.
Case 2: $n_0=1$. Then, for $g(x):=f(x)/x$, the multiple roots of $f$ are the same as those of $g$ (except for the zero-probability event $\{u_1=0\}$) -- which reduces the consideration to
Case 3: $n_0=0$. Then $f(x)=v_0+h_v(x)$, where $h_v(x):=v_1 x^{n_1}+\dots+v_k x^{n_k}$, $v:=(v_1,\dots,v_k)$, and $v_j:=u_{n_j}$ for $j=0,\dots,k$. The random finite set
$$X_v:=\{x\in\mathbb R\colon f'(x)=0\}=\{x\in\mathbb R\colon h'_v(x)=0\}$$
depends only on $v$, but not on $v_0$.
Next, $f$ has multiple roots iff $-v_0$ is in the random finite set $h_v(X_v)=\{h_v(x)\colon x\in X_v\}$. Therefore and because the distribution of $v_0$ is continuous and because $v_0$ is independent of $v$, we conclude that the probability that $f$ has multiple roots is $$P(-v_0\in h_v(X_v))=EP(-v_0\in h_v(X_v)|v)=E0=0.$$
