Clearly, $f$ is real valued, non-negative, and, by Cauchy-Schwarz,
$$f(x, y) = x \cdot \sqrt{1 - y^2} + \sqrt{1 - x^2} \cdot y \le \sqrt{x^2 + (1 - x^2)} \cdot \sqrt{(1 - y^2) + y^2} = 1,$$
so that $f$ maps $S$ into $[0, 1]$. But if $0 \le z \le 1$, then $(z, 0) \in S$ and $f(z, 0) = z$, and so $f$ assumes every value in $[0, 1]$. Hence $f$ is also onto (surjective).

For $x, y \in [0, 1]$ there are unique $\alpha, \beta \in [0, \frac{\pi}{2}]$ such that $x = \sin \alpha$ and $y = \sin \beta$. Because $\cos \alpha$ and $\cos \beta$ are non-negative for such $\alpha, \beta$, we have $\sqrt{1-x^2} = \cos \alpha$ and $\sqrt{1-y^2} = \cos \beta$. Therefore,
$$f(x, y) = x\sqrt{1-y^2} + y\sqrt{1-x^2} = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \sin(\alpha + \beta).$$
For $\alpha, \beta \in [0, \frac{\pi}{2}]$ we have $\alpha + \beta \in [0, \pi]$ and so $\sin(\alpha + \beta) \in [0, 1]$ so that $f$ maps $S$ into $[0, 1]$. Surjectivity follows from $f(z, 0) = z$.
Clearly, $f$ is not one-to-one (injective) since $f(x, y) = f(y, x)$, for example $f(1, 0) = f(0, 1) = 1$.