Each term in the sequence is the maximum number of repeating blocks up to that term. It starts $$1,1,2,1,1,2,2,2,3,1,1,2,1,1,2,2,2,3,2,1,...$$ That is to say, when we have $1,1,2,1,1,2$ the next term is 2 as is shown in the following demonstration $$(1,1,2,1,1,2)=()+(1,1,2)^2 \rightarrow 2$$ Similarly, when the sequence is $1,1,2,1,1,2,2,2$ the next term is a $3$ because $$(1,1,2,1,1,2,2,2)=(1,1,2,1,1)+(2)^3\rightarrow 3$$ Oeis says that the first $4$ in the sequence occurs at the $220$th place. But when I try calculating this, the result is the $23482$th place - which isn't $220$.
Here's my approach. Start by assigning $$a=1,1,2,1,1,2,2,2,3$$ so that we can continue the sequence in a compact way ($a,a,2,...$) and assign a variable to the next chunk
$$b=a,a,2,a,a,2,2,2,3$$ And we repeat this process until a $4$ shows up $$c=b,2,b,2,2,2,3,3$$ $$d=c,2,c,2,2,2,3$$ $$e=d,d,2,d,d,2,2,2,3$$ $$f=e,2,e,2,2,2,3,3$$ $$g=f,2,f,2,2,2,3$$ $$h=g,g,2,g,g,2,2,2,3$$ $$i=h,2,h,2,2,2,3,3,3,4$$ I can see that the sequence online is different from mine even as early as $c$.
How can we predict Gijswijt's sequence quickly and correctly - at least up to the first "$4$"?
The "process" you're repeating is not correct.
The definition of "Gijswijt's sequence" $(a_1a_2a_3...)$ is as follows: $$\begin{align}a_1&=1\\a_{n+1}&=\mathrm{curl}(a_1a_2...a_n),\ \ n\ge 1\end{align}$$ where $\mathrm{curl}(a_1a_2...a_n)$ is the largest $k$ such that $$a_1...a_n=X\,\underbrace{Y\,Y\,...\,Y}_{k\,Y\text{s}}$$ for some strings $X$ and $Y,$ where $X$ may be the empty string, but $Y$ must be non-empty. $(\mathrm{curl}(a_1a_2...a_n)$ is called the "curling number" of $a_1...a_n.)$
Comparing your $a,b,c$ against Gijswijt's sequence, we see that your procedure goes wrong in the $43$rd term:$$\begin{align}a&=112112223 \tag{okay}\\ b=aa2aa2223&=11211222311211222321121122231121122232223\tag{okay}\\ c=b2\color{red}{b}...&=112112223112112223211211222311211222322232\color{red}{1}...\tag{wrong}\\ \\ \text{Gijswjt's seq.}&=112112223112112223211211222311211222322232\color{blue}{2}...\\ &=1121122231121122232112112223112112\ \underline{\color{green}{2232}}\ \underline{\color{green}{2232}}\ \color{blue}{2}... \end{align}$$ Here $a_{43}=2$ (not $1$) because $$a_1...a_{42} = \underbrace{1121122231121122232112112223112112}_{X}\ \underbrace{2232}_{Y}\ \underbrace{2232}_{Y}.$$
As far as I know, the only algorithms for "Gijswijt's sequence" directly search for curling numbers (although maybe not by that name). Unfortunately, such searching will encounter $Y$-blocks of length up to $19$ before the first $4$ appears.
For example, Gijswijt's Python program for $\mathrm{curl}$ (which is posted at OEIS A091579) is easily used to compute the first $220$ terms (where I've boxed the first occurrence of $1,2,3,4$): $$\begin{align}&\boxed{\mathbb{1}}1\boxed{\mathbb{2}}11222\boxed{\mathbb{3}}112112223211211222311211222322232223321121122231121122232112112223112\\&112223222322233222321121122231121122232112112223112112223222322233211211222311\\&211222321121122231121122232223222332223222322233222322232223333\color{blue}{\boxed{\mathbb{4}}}...\end{align}$$ These show a recursive structure as illustrated in the following picture:
However, this structure apparently cannot be used to bypass the computation of curling numbers. That is, although there are patterns in the repeating blocks, there is an "unpatterned" appearance of the terms I've highlighted in yellow (these are the $S_n$ in the following bullet). Although these terms have some structure, they themselves apparently can only be found via curling numbers.
This article proves, among other things, the following:$$\begin{align}a_1a_2a_3... &= \lim_{n\to\infty} B_n\\ B_1 &= 1\\ B_{n+1} &= B_n\,B_n\,S_n,\ \ n\ge 1 \end{align}$$ where the $S_n$ strings (highlighted in yellow above) contain no $1$, and for every positive integer $i$, some $S_n$ contains $i$ (hence the sequence is unbounded).
In the above picture, the first (smallest) box contains $B_1$, the second box contains $B_2$, etc., up to the seventh box, which contains $B_7$ (ending in the first occurrence of $4$).