Due to symmetry we need to define $f(x)=x^2+2$ on $(-1,0)$. It remains to discuss the $2$-periodicity. One starts with "special points", for example $x=0$. Then
$$f(0)=0^2+2=\text{(2-periodicity)}=f(0+2)=f(2),$$
i.e. $f(2):=2$. To find the extension of $f$ on $(2,3)$ one continues with all $\delta\in (0,1)$, i.e.
$$f(\delta)=\delta^2+2=\text{(2-periodicity)}=f(\delta+2),$$
where $\delta+2\in (2,3)$. Similarly, considering all points $-\delta\in(-1,0)$, using $2$-periodicity one finds the extension of $f$ on the interval $(1,2)$ via $-\delta+2\in (1,2)$, as we did above. The extension of $f$ for all other points easily follow by drawing.
Due to symmetry we need to define $f(x)=-x^2-2$ on $(-1,0)$. One selects once again
the "special points", for example $x=0$ as in the even case. The analysis is completely similar, with due changes.
For any irrational $\alpha$, the function $f(x)=\sin x+\sin(\alpha x)$ is not periodic, because $\limsup f(x)=2$ but $f(x)<2$ for every $x\in\Bbb R$.
For a proof, see this and have in mind that $f$ is continuous.
Best Answer
What is meant here is that any $2\pi$-periodic function $f : \Bbb{R} \to \Bbb{C}$ yields a well-defined function
$$ g : S^1 \to \Bbb{C}, e^{i x} \mapsto f(x). $$
Note that this is well-defined, because $e^{i x } = e^{i y}$ implies $x-y \in 2\pi \Bbb{Z}$ and hence $f(x) = f(y)$, because $f$ is $2\pi$-periodic.
Conversely, any function $g : S^1 \to \Bbb{C}$ yields a $2\pi$-periodic function
$$ f : \Bbb{R} \to \Bbb{C}, x \mapsto g(e^{i x}). $$
Finally, you should verify that the two "transformations" $f \mapsto g$ and $g \mapsto f$ defined above are inverse to each other.
Hence, we have constructed a natural identification between both classes of functions.