I have some clarificational questions to this answer (definitions of $F,U,\varepsilon$ should be clear from this):
As to $\varepsilon_V$, you know that the result is an element of $V$ and $\varepsilon_V$, being an arrow in $\mathbf{Vect}_k$ is a vector space homomorphism, so, by definition, it takes the addition (and other operations) of $FU(V)$, which is a "formal" addition, and maps it to the addition of $V$, which is a "real" addition. Of course, $V$ could itself be a vector space of formal linear combinations, but the point is that $\varepsilon_V$ has no choice but to interpret the "formal" operations in $FU(V)$ as the operation so $V$ whatever they are. Formal $k$-linear combinations are not, in general, elements of $V$.
From the same answer, formal linear combinations are of the form $r_{s_1}\cdot v_{s_1}+\cdots+r_{s_n}\cdot v_{s_n}$ where $s_1,\dots,s_n\in S$ and are distinct, and $r_{s_1},\dots,r_{s_n}\in k$ (in our case, $S=U(V)$). Consider the simplest case when the formal linear combination is just $v_{s_1}$. There's no any formal addition or multiplication. Is this formal combination supposed to be mapped to $s_1$ under $\varepsilon_V$? (I believe the answer I referred to above doesn't discuss this.) Further, is $r_{s_1}\cdot v_{s_1}$ supposed to be mapped to $r_{s_1}\cdot_V {s_1}$, where $\cdot $ is a formal multiplication and $\cdot_V$ is mupltiplication $V\times k\to V$? And then more generally, I suppose $$r_{s_1}\cdot v_{s_1}+\cdots+r_{s_n}\cdot v_{s_n}\mapsto r_{s_1}\cdot_V s_1 +_V\dots+_V r_{s_n}\cdot_V s_n$$ where $+_V$ is addition on $V$ (which is, I believe, referred to as "real" addition in the quote above). Is that right?
And lastly, how do we know that $\varepsilon: FU(V)\to V$ does exactly what the answer I referred to described (and what I tried to expand on)? Is it because this is the only sensible way to define $\varepsilon_V$ and because the counit of adjunction is unique? Or does this follow from some facts in the theory of "term algebras" mentioned in the answer?
To your first question: Yes, you are right. In the answer you are referring to, the notation $v_s$ instead of $s$ was used to better distinguish the formal sum $\sum r_s v_s$ from the sum taken in $V$, $\sum r_s s$. Under the counit $\epsilon$, $v_s$ is then mapped to $s$, and the formal sum $r_{s_1}v_{s_1}+\dots +r_{s_n}v_{s_n}$ is mapped to the evaluation of this sum in $V$, which is $r_{s_1}s_1 + \dots r_{s_n}s_n$.
To the second part: Look at the more general setting of an adjunction
Let $F: C\rightarrow D$ and $G: D\rightarrow C$ be functors together with a natural isomorphism with components $D(Fc,d) \simeq C(c,Gd)$. Then,fixing $c$, one can see that the object $Fc$ represents the functor $d\mapsto C(c,Gd)$. By the Yoneda lemma, one can see that the natural isomorphism $D(Fc,-) \Rightarrow C(c,G-)$ is determined by an element of $C(c,GFc)$, which we call $\eta_c$ and which we call the component of the unit of the adjunction on $c$. In the same manner, the observation that $Gd$ represents the contravariant functor $c\mapsto D(Fc,d)$ gives an element $\epsilon_d$ of $D(FGd,d)$, which we call component of the counit at $d$.
This discussion shows that the counit is obtained by the construction of the Yoneda lemma: If we call the natural isomorphism of the adjunction $\alpha$, then $\epsilon_d$ is given by $\alpha^{-1}_{Gd,d} \text{Id}_{Gd}$, which is often called the transpose of $\text{Id}_{Gd}$.
So to find $\epsilon_V$ in this particular case, we have to find the transpose of $\text{Id}_{UV}$ under the free-forgetful adjunction of $\text{Vect}_k$. So we first need to know: if we are given a set $S$ and a vector space $V$, which morphism in $\text{Vect}_k$ does a morphism $S\rightarrow UV$ correspond to under this adjunction? The answer is the following: A map $f: S\rightarrow UV$ is send to the vector space homomorphism $\phi: FS \rightarrow V$ that sends $s\in S$ to the vector $f(s)$, but this time seen as element of $V$ instead of $UV$, and a formal sum $\sum r_{s_i} v_{s_i}$ in $FS$ to the sum in $V$ that is $\sum r_{s_i} f(s_i)$.
So, what linear map does the identity morphism in Set $\text{Id}: UV\rightarrow UV$ correspond to? Well, a formal sum in $UV$ is exactly sent to its result in $V$ under the vector space addition and scalar multiplication!
Therefore, the counit of the free-forgetful adjunction works as described in the question you linked.