Continued fraction for $c= \sum_{k=0}^\infty \frac 1{2^{2^k}} $ - is there a systematic expression?

Shallit's paper, which you've cited, gives a simple algorithm for generating these coefficients. It works for $\sum_{k=0}^{\infty}u^{-2^k}$ for integer $u\ge 3$; but he notes that it also works for $u=2$ (your case) with a slight modification.

Start with $$B_1=[1,3].$$ Then repeatedly apply the following rule: $B_{n+1}$ is generated from $B_n$ by appending the reverse of $B_n$ to $B_n$ and then adding $1$ and $-1$ to the two central terms. That is, $$ B_2=[1,4,2,1] \\ B_3=[1,4,2,2,0,2,4,1] \\ B_4=[1,4,2,2,0,2,4,2,0,4,2,0,2,2,4,1] \\ ... $$ This is correct as it stands, but generally one excludes zeroes from the entries of a continued fraction. To rectify that (this is the slight modification), generate $C_n$ from $B_n$ by contracting any subsequence $[a,0,b]$ down to $[a+b]$. That is, $$ C_2=[1,4,2,1] \\ C_3=[1,4,2,{\mathbf{4}},4,1] \\ C_4=[1,4,2,{\mathbf{4}},4,{\mathbf{6}},{\mathbf{4}},2,4,1]\\ ... $$ Now the elements of each $C_n$ (except the final $1$) are the entries in the continued fraction for $\sum_{k=0}^{\infty}2^{-2^{k}}$. (You'll note that they agree with the ones you've already found.) Just choose a large enough $n$ to get to the entry you need.