Can any smooth hyperelliptic curve be embedded in a quadric surface? Let's consider algebraic curves over a fixed algebraically closed field $K$.
It's well known, that every smooth elliptic curve (genus $g = 1$) can be embedded in a quadric surface in $\mathbb{P}^3$. This fact follows simply from the Riemann–Roch theorem.
More generally, for smooth hyperelliptic curves of higher genus ($g \ge 2$) it's known that such curves can be embedded in a quadric in weighted projective space $\mathbb{P}(1,1,g)$, see, for example, work of D. Eisenbud. (So, in the case $g=2$ we have embedding in $\mathbb{P}^4$).

But, can any smooth hyperelliptic curve $H$ be embedded in a quadric surface in $\mathbb{P}^3$?

It's natural question, because, by the definition we have mophfism $\phi:H \to \mathbb{P}^1$ of degree 2.
I think, this problem is connected with topics "Families of hyperelliptic curves and double covers of quadric surface" and Quotient Surface of A Hyperelliptic Involution, but I don't get it. (I'm interested in the case of any algebraicaly closed field and in the case of finite field).
 A: The answer is yes in the case $g=2$.
In other words,  any curve $C$ of genus $2$ can be embedded in a quadric $Q \subset \mathbb{P}^3$. In fact, let us consider a divisor $D$ of degree $5$ on $C$. Then $D$ is very ample [Hartshorne, Corollary 3.2 page 308] and $h^0(D)=4$, so it defines an embedding
$\phi \colon C \to \mathbb{P}^3$ such that its image (that we call again $C$) is a curve of degree $5$.
Now we use Castelnuovo theory: if $C$ is any non-degenerate curve  of genus $g$ and of odd degree $d$ in $\mathbb{P}^3$, then $$g \leq \frac{1}{4}(d^2-1)-d+1,$$ and if equality holds then $C$ lies on a quadric surface [Hartshorne, Thm. 6.4 page 351]. 
Since equality does hold for $(g, d)=(2,5)$, we are done.
ADDENDUM. Just for completeness, let me add a deformation theory argument showing that the general smooth hyperelliptic curve of genus $g$ lies on a quadric $Q$. This will replace my previous imprecise naive count of parameters. Notice that Jack Huizenga showed in his answer that this is actually true for any such a curve.
Let us start with a curve $C \subset Q$ of bidegree $(2, g+1)$. This is clearly hyperelliptic and we have a short exact sequence $$0 \to T_C \to T_Q \otimes \mathcal{O}_C \to N_{C/Q} \to 0$$
and since $H^0(T_C)=0$ for $g \geq 2$ this gives in turn a sequence in cohomology
 $$0 \to H^0(T_Q \otimes \mathcal{O}_C) \to H^0(N_{C/Q}) \stackrel{\delta}{\to} H^1(T_C).$$
Now standard computations yield
$$h^0(T_Q \otimes \mathcal{O}_C)=g+6, \quad h^0(N_{C/Q})=3g+5$$
hence the image of the map $\delta \colon H^0(N_{C/Q}) \to H^1(T_C)$ has dimension $3g+5-g-6=2g-1.$
This means that the embedded deformations of $C$ in $Q$, which are all hyperelliptic,  form a family of dimension $2g-1$. But this is exactly the dimension of the hyperelliptic locus in $\mathcal{M}_g$, since any hyperelliptic curve of genus $g$ is a double cover of $\mathbb{P}^1$ branched in $2g+2$ points.
A: Yes.
Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$.  By Riemann-Roch, $\dim|L| = 1$ and $|L|$ is base-point free, 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.  The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution.  But in this case the assumption that $\dim |L|=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points.  Since $L$ was general, this isn't true, and we're done.
Here's a cute (although trivial) kind of partial converse:  If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.
A: An explicit realization of degree $2$ and degree $g+1$ maps that separate points can be provided. Suppose the equation of a hyperelliptic curve is $$C:y^2=f(x)$$ with $\deg(f)=2g+2$. "Complete the square" by writing $$f(x)=r(x)^2+q(x)$$ with $\deg(r)=g+1$ and $\deg(q)\leq g$. Then, the maps $$A:(x,y) \mapsto x$$ $$B:(x,y)\mapsto y-r(x)$$ are degree $2$ and degree $g+1$ respectively. The map $B$ is degree $g+1$ because if we assume that $B(x,y)=c$, then we get the equation $c(c+2r(x))=q(x)$, which generically has $g+1$ solutions. Furthermore $(A,B)$ is clearly injective. So, the image $$(A,B):C\mapsto C'\subset\mathbb{P}^1\times\mathbb{P}^1$$ is a degree $(2,g+1)$ curve that $C$ normalizes and the logic of Jack's answer applies.
