For each $g$ there is $[C]\in\mathcal{M}_g$ which embeds on a K3 surface For each genus $g\geq 0$ there is a (smooth irreducible) curve $C$ of genus $g$ which embeds on some K3 surface.
How does this follow from the surjectivity of the period map for K3 surfaces?
Is there a simpler reason for this (apparently) simple fact?
(it seems to me a bit of an overkill to invoke such a strong result - can we actually avoid it?)
 A: First note that if $C$ is a curve on a $K3$ surface $X$, then by adjunction we have
$$g(C) = \frac12 C^2 +1.$$
Now to your question: the cases $g=0$ and $g=1$ (rational and elliptic curves) are easy enough to deal with by hand. So we need to worry about $g \geq 2$. The precise statement about periods (i.e. lattices) we need is

Proposition: For each $d \geq 1$, there exists a K3 surface $X$ (actually a 19-dimensional family of them, but who's counting) with an ample line bundle $H$ such that such that
  $$\operatorname{Pic}(X) = \mathbf Z \cdot H \ \text{ with } H^2= 2d.$$

(This follows from surjectivity of the period map.)
Now it is apparently a well-known fact (full disclosure: I didn't know it until today!) that every ample linear system on a K3 surface is basepoint-free. Hence by Bertini, the linear system $|H|$ corresponding to the line bundle $H$ above  contains a smooth curve $C$. Putting $C^2=2d$ in the formula above, we get the statement you want.
Update: As my close associate Nefertiti mentioned in a comment above, there are more elementary constructions of K3 surfaces with ample line bundles of self-intersection $2d$. See Theorem 2.4.2 (and the references there) on p.35 of Huybrechts' notes.
