Is Functional Complete means Turing Complete?
A Turing machine defines a function over all of $\Sigma^*$, it is a finite object which describes the operation on infinitely many inputs.

A logical circuit however, is defined over a specific number of variables, so it has a finite domain. Using boolean circuits over $n$ variables we can implement the set of all functions from $\left\{0,1\right\}^n$ to $\left\{0,1\right\}$, and no more (so no Turing completeness in the simple interpretation).

This is why, when dealing with logical circuits, the notion of non uniform computation is introduced. By non uniform, we mean that the mechanisem for processing some input $x$ might depend on $|x|$. In the context of circuits, this means that instead of talking about one circuit, were talking about a circuit family $\left\{c_n\right\}_{n\in\mathbb{N}}$, where $c_n$ is a circuit with $n$ inputs.

We say that a circuit family $\mathcal{C}_n$ decides some language $L$ if $\forall x\in\Sigma^*: x\in L \iff c_{|x|}(x)=1$. Note that non uniform computation extends the classical uniform computation, i.e. using circuit families of arbitrary size we can decide undecidable (relative to Turing machines) languages. Actually, we can decide all languages. If you want to reduce this back to the Turing model, then you must require the existence of a computable function $f$ which, given $n$, generates the circuit $c_n$.