Think of starting at $\langle 0,0\rangle$ and following a path determined by the binary string, taking one step for each bit: if you’re at $\langle m,n\rangle$, go to $\langle m+1,n+1\rangle$ if the next bit is $0$, and to $\langle m+1,n-1\rangle$ if it’s $1$. (In other words, a $0$ bit indicates a step up and to the right, while a $1$ bit indicates a step down and to the right.) The condition on substrings amounts to saying that if $h$ and $\ell$ are the maximum and minimum $y$-coordinates along the path, then $h-\ell\le 2$. Thus, every allowable path lies between $y=0$ and $y=2$, between $y=-1$ and $y=1$, or between $y=-2$ and $y=0$. Call $h-\ell$ the width of the path.
Suppose that an allowable path of length $n$ terminates at $\langle n,2\rangle$; the entire path must lie between $y=0$ and $y=2$, so it’s easy to see that we can extend the bit string for the path by $10$ or by $11$ and still have an allowable path.
Show that no matter where a path of length $n$ and width $2$ terminates, there are exactly $2$ two-bit extensions of its bit string that yield allowable paths.
Show that there are exactly two allowable paths of length $n$ and width $1$, and that each of these produces $3$ allowable paths of length $n+2$, two of which are of width $1$.
Conclude that if $a(n)$ is the number of allowable paths of length $n$, then $$a(n+2)=2a(n)+2\;.$$
You know $a(0)$ and $a(1)$, so you can solve this as two separate recurrences, one for even $n$ and one for odd $n$. You might even want explicitly to set $b(n)=a(2n)$ and $c(n)=a(2n+1)$ for $n\in\Bbb N$, so that $b(n+1)=2b(n)+2$ and $c(n+1)=2c(n)+2$; these sequences satisfy the same recurrence, so you need only solve it once in general and then substitute in the different initial values.