Trace ideal of a projective module In his 1969 paper "On projective modules of finite rank", Wolmer Vasconcelos writes

Let $M$ be a projective $R$-module... The trace of $M$ is defined to be the image of the map $M \otimes_R \operatorname{Hom}_R(M, R) \to R$, $m \otimes f \to f(m)$; it is denoted by $\tau_R(M)$. If $M \oplus N = F$ (free), it is clear that $\tau_R(M)$ is the ideal of $R$ generated by the coordinates of all elements in $M$, for any basis chosen in $F$. It follows that for any homomorphism $R \to S$, $\tau_S(M \otimes_R S) = \tau_R(M) S$.

A similar claim appears in his 1973 paper "Finiteness in projective ideals":

We recall the notion of trace of a projective module $E$ over the commutative ring $A$. It is simply the ideal $J(E) = J = \Sigma f(E)$ where $f$ runs over $\operatorname{Hom}_A(E, A)$. Equivalently, $J$ is the ideal generated by the “coordinates” of all the elements of $E$ whenever a decomposition $E \oplus G = F$ (free) is given. Under the second interpretation, it follows that if $h \colon A \to B$ is a ring homomorphism, then $J(E \otimes_A B) = h(J(E)) B$.

The first claim is easy to verify (albeit with a change of free module) as follows. If $F = M \oplus N$ has a basis $\{ v_i \}$, we can consider $F'  = F \oplus R$, where $u$ is a generator for $R$. For any $f \colon F \to R$, $F'$ has a basis composed of $u$ and all $w_i := v_i - f(v_i)u$. With respect to this basis, $v_i = w_i + f(v_i) u$, hence the $u$-coordinate of $v_i$ is $f(v_i)$. So every homomorphism $M \to R$ is the restriction of a coordinate function on $F'$.
What is not clear to me is the reason for the second claim that $\tau_S(M \otimes_R S) = \tau_R(M) S$. The inclusion $\tau_R(M) S \subset \tau_S(M \otimes_R S)$ is obvious, so let me focus on the other one.
One can choose a decomposition $M \oplus N = F$ (free), so that $M \otimes_R S \oplus N \otimes_R S = F \otimes_R S =: F_S$, which is free over $S$. Up to adding a $S$ summand, one can also assume that every $f \colon M \otimes_R S \to S$ is the restriction of some coordinate function on $F_S$. But the coordinates on $F_S$ depend on the choice of a basis. If the basis is obtained from an $R$-basis of $F$, the claim is clear. But $F_S$ could have many choices of $S$-bases which are not derived from $R$.
I think I am missing something quite trivial, but I cannot see it right now, so I though I'd rather ask here.
 A: The confusion is linguistic, as identified in the comments.
Lemma. Let $M$ be a projective $R$-module, and suppose $M \oplus N \cong F$ is free on a basis $\mathcal B$. For $b \in \mathcal B$, write $\varepsilon_b \colon F \to R$ for the 'dual' element taking $b$ to $1$ and all other basis elements to $0$. Then $\tau(M)$ is the ideal generated by $\varepsilon_b(m)$ for $b \in \mathcal B$ and $m \in M$.
(By abuse of notation, we write $\varepsilon_b(m)$ for what should properly be denoted $\varepsilon_b(m,0)$.)
Proof. Since $\varepsilon_b|_M$ is a homomorphism $M \to R$, we clearly have $\varepsilon_b(m) \in \tau(M)$ for all $b \in \mathcal B$ and all $m \in M$. We have to show that they generate. In the definition of $\tau(M)$, we may replace $\operatorname{Hom}(M,R)$ by $R^{\mathcal B} = \operatorname{Hom}(F,R) \twoheadrightarrow \operatorname{Hom}(M,R)$. Elements can be written as $f = (f_b)_{b \in \mathcal B}$, where $f_b = f(b)$ are constants. Now the idea is that $f(m)$ only depends on the coordinates of $f$ where $m$ is supported:
Let $f = (f_b)_{b \in \mathcal B} \in R^{\mathcal B}$ and $m = \sum_{b \in \mathcal B'} a_b b \in M$ for some finite subset $\mathcal B' \subseteq \mathcal B$. Write $f_{\mathcal B'}$ for the function whose $\mathcal B'$-coordinates agree with $f$ and whose other coordinates vanish. Then
$$f(m) = \sum_{b \in \mathcal B'} f(a_b b) = \sum_{b \in \mathcal B'} f_b \cdot a_b = \sum_{b \in \mathcal B'} f_b \cdot \varepsilon_b(m),$$
so $f(m)$ is expressed as a combination of the $\varepsilon_b(m)$. $\square$
Corollary. Let $M$ be a projective $R$-module, and let $R \to S$ be a ring homomorphism. Then
$$\tau\left(M \underset R\otimes S\right) = \tau(M)S.$$
Proof. Write $M \oplus N \cong F$ for some $R$-module $N$ and a free $R$-module $F$. Then
$$\left(M \underset R\otimes S\right) \oplus \left(N \underset R\otimes S\right) \cong F \underset R\otimes S.$$
If $F$ has basis $\mathcal B$, then the elements $b \otimes 1$ form a basis of $F \otimes_R S$. Moreover, $M \otimes_R S$ is generated by elements of the form $m \otimes 1$. Therefore, $\tau(M \otimes_R S)$ is exactly the ideal generated by $\varepsilon_{b \otimes 1}(m \otimes 1)$, which is $\tau(M)S$. $\square$
