In general **no such stratification exists**. Lipschitz functions coincide with $C^1$ functions on sets of positive measure: 

**Theorem.** *If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ there is $g\in C^1(\mathbb{R}^n)$ such that the Lebesgue measure of the set 
$\{ f\neq g\}$ is less than $\epsilon$.* 

You can find a proof of this result in almost any book on geometric measure theory, see for example Theorem 1 in Section 6.6.1 in  L. C. Evans, R. F. Gariepy, *Measure theory and fine properties of functions*. Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992.

Therefore, you can decompose the graph of $f$ into countably many $C^1$ pieces (defined on measurable sets) and a set of measure zero.  However, in general a Lipschitz function need not coincide with a $C^2$ function on a set of positive measure so there is no hope for any sort of $C^2$ stratification even for functions defined on a real line. 

By the way, I checked the book by  Vladimir Shikhman and there is no such result as the one formulated in the question. Instead, they *consider Lipschitz functions $f : \mathbb{R}^n\to \mathbb{R}$ whose graphs admit a $C^\infty$-
Whitney stratification*, and that is a different story. No theorem, but assumption.