If $(R,\mathfrak m)$ is a local domain which is not a field, then for its fraction field $K$ we have $\mathfrak m K=K$.
Amusingly, this proves that $K$ is not a finitely generated $R$-module.
You are correctly interpreting Lang's induction: he is arguing that if $\overline{x}_1,\dots,\overline{x}_r$ are linearly independent over $A/\mathfrak{m}$, then $x_1,\dots,x_r$ are linearly independent over $A$ in $E$. Karl Kronenfeld sketched an answer to the question in comments, but I'll write out the details.
Linear independence of $\overline{x}_1$ over $A/\mathfrak{m}$ is equivalent to the statement that $x_1\notin \mathfrak{m}E$. In particular, it is not zero. As Karl mentioned, Lang needs this in order to conclude that $ax_1=0$ implies $a=0$.
The other piece of the picture (as again Karl mentioned) is the implicit use of Nakayama's lemma. Nakayama's lemma states that for any finitely generated module $M$ over $A$, $\mathfrak{m}M=M$ cannot hold unless $M=0$. In the present situation, this will be applied with $M=E/E'$, where $E'$ is the submodule of $E$ generated by $x_1,\dots,x_n$, where $\overline{x}_1,\dots,\overline{x}_n$ are a basis for $E/\mathfrak{m}E$ over $A/\mathfrak{m}$, as follows:
Since $x_1,\dots,x_n$ span $E$ mod $\mathfrak{m}E$, we have $E = E' + \mathfrak{m}E$, and it follows that any element of $E$ differs from an element of $\mathfrak{m}E$ by an element of $E'$, i.e. that every element of $E$ is actually equal to an element of $\mathfrak{m}E$ mod $E'$, i.e. that $\mathfrak{m}(E/E') = E/E'$. Since $E$ is finitely generated, so is $E/E'$, and Nakayama's lemma then allows us to conclude that $E/E'=0$, i.e. that $E=E'$, i.e. that $x_1,\dots,x_n$ span $E$. (This argument uses only $E$'s finite generation and makes no use of the fact that it is projective.)
Now we can apply the proof you quote, to conclude that (now quite heavily using $E$'s projectivity), $\overline{x}_i$'s linearly independent over $A/\mathfrak{m}$ lift to $x_i$'s linearly independent over $A$. Thus a basis lifts to a basis, and we get to conclude that $E$ is free.
Best Answer
Let $K=k(t)$, $W_1 = K \otimes_S R,M_1=R$. Then $W_1$ is a finite dimensional $K$-vector space and $M_1$ is a finitely generated sub-$S$-module of it, where $S=k[t]$ is a PID.
Take $m_1\ne 0\in M_1$, then $K m_1\cap M_1 = S b_1$ for some $b_1\in M_1$ (this is where we need that $S$ is a PID and that $M_1$ is finitely generated). Let $$M_2 = M_1/Sb_1, \qquad W_2 = W_1/Kb_1$$
And then repeat with $M_2,W_2$ which satisfy the same hypothesis, obtaining $b_2$ and then $M_3,W_3$, and so on.
$\dim_K(W_n)$ decreases at each step so the process terminates, and the obtained $b_1,\ldots,b_d$ are a free $S$-basis of $M_1=R$.
(take any representative in $M_1$ of $b_2$ which at first is in $M_1/Sb_1$)