Disclaimer: I am not an expert myself so my approach might be not the easiest or wrong somewhere. (But I think it's fine.)
The intuitive part is that for fibrant objects, $\pi(\varinjlim Y_i)=\varinjlim \pi(Y_i).$ This is because for these objects homotopy groups are classes of maps. Recall how it goes, for example, for CW-complexes. Any map $S^k \to X =\varinjlim X^n$ falls into some $X^n$ and thus $\varinjlim \pi(X^n) \to \pi(X)$ is surjective. Likewise, any homotopy witnessing that $[\alpha]=[\beta] \in \pi(X)$ factors through some finite-stage $X^n$ and so our map is also injective. Here it works in just the same way using that, as Goerss-Jardine puts it on p.103, $\partial\Delta$, and also cylinders, are small, which means formally that functors they represent commute with filtered colimits, and informally that maps from them factor through some "finite-stage" elements of the colimit diagram. It follows that the map $\varinjlim \pi(Y_i) \to \pi(\varinjlim Y_i)$ is surjective and injective.
Now, one way to compute $\pi(X)$ in general is to replace $X$ with a fibrant object $Y$ and compute $\pi(Y)$ as homotopy classes of maps. In particular, Kan's $\text{Ex}^\infty$ works for that (e.g. Goerss-Jardine III.4, also answers here, and properties listed here). $\text{Ex}^\infty$ is a way to get from an object a fibrant object weakly equivalent to it. This functor commutes with filtered colimits. Then we have $\pi(\varinjlim X_i)=\pi(\text{Ex}^\infty\varinjlim X_i)=\pi(\varinjlim \text{Ex}^\infty X_i)=\varinjlim\pi( \text{Ex}^\infty X_i),$ the last equality following from the previous paragraph.
