According to the text leading up to exercise 8 (in Seven Sketches), \$$\textrm{Ob}(\textbf{Free}(G))=V\$$, and the morphisms are paths (i.e., possibly infinite lists of arrows \$$a,b,c\ldots{}\in A\$$ such that \$$t(a)=s(b)\$$, \$$t(b)=s(c)\ldots\$$), including paths of length zero, i.e., identity arrows. Where \$$V\$$ is the set of vertices of the graph, \$$A\$$ is the set of arrows (directed edges, including loops from a vertex to itself), and \$$s,t:A\to V\$$ are the source and target maps.

a) Composing an arrow \$$f:c\to d\$$ with the identity on either side yields a path of the same length, one, containing equivalent arrows in the same order. Since paths are lists of arrows with matching sources and targets, where identity arrows can be omitted, \$$\mathrm{id}_d\circ f\$$, \$$f\$$, and \$$f\circ\mathrm{id}_c\$$ all yield the same/equivalent path(s), and so the morphisms are equal.

b) Since paths, i.e., lists of arrows, are associative (i.e., \$$(ab)c=a(bc)\$$), morphisms are also. This is because morphisms are equivalence classes of paths, where paths with or without identity arrows are considered equivalent (and one makes only this identification), and associativity of paths is respected by this equivalence (e.g., \$$(h\circ g)\circ (f\circ \mathrm{id}\_{c_0})\$$ \$$=(h\circ g)\circ f=h\circ (g\circ f)\$$ yields the same thing as \$$(h\circ g)\circ (f\circ \mathrm{id}\_{ c_0})\$$ \$$=h\circ (g\circ(f\circ\mathrm{id}\_{c_0}))\$$ \$$=h\circ (g\circ f\$$).

Edit: This can be seen as a consequence of the commutativity of a certain diagram corresponding to the equation: \$$\alpha\circ\sim=\sim\circ\alpha\$$, where \$$\alpha\$$ serves to switch the position of parentheses and \$$\sim\$$ identifies paths differing only by identity arrows.