You simply have that every $0$ must be followed by a $2$, so $s_n$ is the number of arrangements of $a$ blocks of type "$02$", $b$ blocks of type "$1$" and $c$ blocks of type "$2$" such that $2a+b+c=n$. On the other hand, every admissible string starts with a "1", then the tail is counted in $s_{n-1}$, with a $2$, then the tail is counted in $s_{n-1}$, or with a $02$, then the tail is counted in $s_{n-2}$, and recurrence relation is $s_n=2s_{n-1}+s_{n-2}$, with $s_1=2$ and $s_2=5$. The recurrence relation and the initial conditions give:
$$ s_n = c_0(1+\sqrt{2})^n+c_1(1-\sqrt{2})^n, $$
$$c_0(1+\sqrt{2})+c_1(1-\sqrt{2}) = 2,$$
$$c_0(3+2\sqrt{2})+c_1(3-2\sqrt{2})=5,$$
$$ c_0-c_1=\frac{\sqrt{2}}{2},$$
$$ c_0+c_1=1, $$
$$ c_0 = \frac{2+\sqrt{2}}{4},\qquad c_1=\frac{2-\sqrt{2}}{4}$$
$$ s_n = \frac{1}{2}\left(p_{n+1} + p_{n}\right), \qquad p_n = \frac{1}{2}\left((1+\sqrt{2})^n+(1-\sqrt{2})^n\right),$$
where $p_n$ is half the Pell-Lucas number $Q_n$. In a combinatorial flavour, you can count the arrangements of $a$ blocks of type "$02$" (with $a\leq\frac{n}{2}$) in $n$ positions times $2^{n-2a}$ (the number of ways to fill the remaining part of the string with $1$s and $2$s). The number
of such arrangements is equal to the number of ways to write $n-2a$ as a sum of $a+1$ natural numbers, so the coefficent of the monomial $x^{n-2a}$ in $(1+x+x^2+\ldots)^{a+1}=\frac{1}{(1-x)^{a+1}}$, that is $\binom{n-a}{a}$.
This gives:
$$ s_n = \sum_{a=0}^{\lfloor n/2\rfloor}\binom{n-a}{a}2^{n-2a}. $$
