How can the VC-dimension of Turing machine be finite?

Question

The VC-dimension of a hypothesis class $\mathcal{H}$ is defined to be the size of the maximal set $C$ such that $\mathcal{H}$ cannot shutter. This paper shows that the VC-dimension of the set of all Turing Machines with $n$ states is $\Theta(n \log n)$.

However, suppose that we take the set of all such Turing machines, for $n$ sufficiently large so that the universal Turing machine is a member of $\mathcal{H}$. The result states that there exists a set $C$ (wlog, $C \subset \{0,1\}^*$) of size, say, $n^2$, such that $\mathcal{H}$ cannot shatter. To my understanding, it means that there exists a function $f : C \rightarrow \{0,1\}$ ("labeling"), such that for every $h \in \mathcal{H}$, it holds that $h \neq f$. Since the elements of $\mathcal{H}$ are Turing machines, I say that "$h$ computes $f$" when the machine $h$ implements $f$.

But $C$ is finite hence $f$ is clearly computable, thus there is some Turing machine $M_C$ which computes it, therefore $M_C$ can be simulated by the universal Turing machine, which is in $\mathcal{H}$, and this is a contradiction (since we assumed $\forall h \in \mathcal{H}, f \neq h$ ). Where is the problem with this argument?