Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $r(L) = 1$, and so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.
To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.
Finally we must see that the map is birational. Let $p$ be a general point of $C$. Then $f(p) = f(q)$ for $q\neq p$ occurs only if $q$ is conjugate to $p$ under the hyperelliptic involution and the divisor of $L$ containing $p$ also contains $q$. As $L$ is general, we conclude that $f(p)\neq f(q)$ for every $q\in C$; thus $f$ is generically injective and hence birational.
Disclaimer: I haven't thought about what happens in characteristic $p$ for the last argument.