Fixed base exponentiation with precomputations
Your question hits at the main part of the faster algorithm presented in the paper. Many people know the usual (base-2) "repeated squaring" algorithm to compute $g^n$ is to write $n$ in binary:
$$
n=\sum_{i=0}^m e_i2^i\qquad e_i\in\{0,1\}
$$
so
$$
g^n=g^{e_02^0+e_12^1+e_22^2+\dotsm}=(g^{2^0})^{e_0}(g^{2^1})^{e_1}(g^{2^2})^{e_2}\dotsm
$$
and observe that $g^{2^k}=(g^{2^{k-1}})^2$ so computation of the $g$ terms can be accomplished by repeated squaring, either in the body of the loop or by precomputation.
The paper simply generalizes this using a base, $b$, other than $2$. If we keep with your example of $b=3$, then we're doing the same thing as the usual algorithm, only using repeated cubing, rather than repeated squaring, so if we have, in general
$$
n=\sum_{i=0}^m e_ib^i\qquad e_i\in\{0,1,\dotsc,b-1\}
$$
then we can write
$$
g^n=(g^{b^0})^{e_0}(g^{b^1})^{e_1}(g^{b^2})^{e_2}\dotsm
$$
and do the usual algorithm, forming the product seriatim, from $0$ up to the limit $m$, using repeated exponentiation of the terms, since $g^{b^k}=(g^{b^{k-1}})^b$ as above. This is your proposal and it will work just fine.

The key idea in the paper relies on the observation that the $e_i$ are now not just $0$ or $1$, so we often have another exponentiation step for each term. Rather than do several of those, the paper just does it once, for each possible value of $e_i$.

Here's your example, with $n=142$ and $b=3$. Since $142_{10}=12021_3$ we'll have $e_4=1$, $e_3=2$, $e_2=0$, $e_1=2$, $e_0=1$. Looping from $i=0$ to $4$ you could successively form the product
$$
g^{142}=(g^{3^0})^1(g^{3^1})^{2}(g^{3^2})^{0}(g^{3^3})^{2}(g^{3^4})^{1}
$$
The paper instead loops first by the nonzero values of the $e_i$, effectively grouping several exponentiations into one, doing this:
$$
g^{142}=\left [(g^{3^1})(g^{3^3})\right ]^2\left [(g^{3^0})(g^{3^4})\right ]^1
$$
obviously saving some multiplications.