Using Friedman's notation, let $\hat{\Delta}$ denote the category of nonempty finite ordinals and strictly increasing maps between them, and $\Delta$ the category of nonempty ordinals and monotonic (i.e. non-decreasing) maps betwen them. Let $i : \hat{\Delta}\to \Delta$ be the inclusion functor. Then a delta set is exactly a contravariant functor $\hat{\Delta}^{op}\to \mathbf{Sets}$, and a simplicial set is exactly a contravariant functor $\Delta^{op}\to\mathbf{Sets}$.
There is an obvious "forgetful" functor from simplicial sets to delta sets, given by forgetting about the degeneracy maps. Another way to describe this would be as follows:
If $X : \Delta^{op}\to \mathbf{Sets}$ is a simplicial set, then $X\circ i : \hat{\Delta}^{op} \to \mathbf{Sets}$ is the corresponding delta set.
Theorem: The functor "precompose with $i$" : $\mathbf{SSet}\to \delta\mathbf{set}$ has both a left and a right adjoint. This theorem comes from the general theory of Kan extensions; you can read about it in Chapter X, Section 3 of "Categories for the Working Mathematician" by Mac Lane. Presumably, here, Friedman is talking about the left adjoint, which we can see as the 'free' simplicial set generated by the underlying delta set.
Let $X$ be a delta set. I will sketch the construction of the free simplicial set $L(X)$ that it generates. If we simply apply the general formula for the left Kan extension, on the ordinal $[n]$, then $L(X)([n])$ can be described as
$$ \operatorname{colim}( (i\downarrow[n]) \xrightarrow{\pi} \hat{\Delta}^{op}\xrightarrow{X}\mathbf{Sets})$$
The above formula is a bit complicated, and all the more so because of the opposite categories involved, so I will unpack it. Let us denote by $T$ the set of all ordered triples $([m], f : [n]\to [m], x)$, where $[m]$ is a finite ordinal, $f$ is a monotonic (i.e., nondecreasing) map from $[n]$ to $[m]$, and $x$ is an element of $X([m])$. We will impose upon $T$ the equivalence relation generated by insisting that $([m],f : [n]\to [m], x)\sim ([m'], f' : [n]\to [m'], x')$ if there is some strictly increasing $g : [m']\to [m]$ such that $g\circ f' = f$ and $X(g)(x) = x'$.
Then we define $L(X)[n]$ as $T/\sim$. I will leave the definition of the behavior of $L(X)$ on morphisms as evident.
The elements of the form $([n], id_{[n]} : [n]\to [n],x)$, we can identify with the original elements of $X([n])$.
Note that whenever one has an element of the form $([m],f : [n]\to [m],x)$, and $f$ is strictly increasing, then this element is $\sim$-equivalent to $([n],id_{[n]}, X(f)(x))$.
It would be a good exercise to look up some of the nice "normal form theorems" for morphisms in the simplicial category, $\Delta$, that state that every morphism can be uniquely factored into face and degeneracy maps in a unique way. You can apply these theorems here to get a nicer description of $L(X)$. For example, given an equivalence class in $L(X)([n])$, can you always choose a representative in $T$ where the morphism consists exclusively of the composition of degeneracy maps and no face maps? This would be nice as it would show that only degeneracies are formally added to $L(X)$, no new faces are created. It is known that elements of a simplicial set can always be written uniquely as some minimal number of degeneracies away from a nondegenerate simplex (see the paper by Milnor "The Geometric Realization Of A Semi-Simplicial Complex" - what effect does this have on choosing representations for $L(X)$?
