Step in Bourbaki Commutative algebra exercise

Here is a self-contained proof, adapted from the proof in Milne's Étale Cohomology referenced in the other answer. We use induction on $n$, the base case $n=1$ being given. Now suppose $n>1$ and the result is known for $n-1$. We can choose $u:F\to K$ such that $u(x_1)=x_1$ and then using the induction hypothesis we can choose $v:F\to K$ such that $v(x_i-u(x_i))=x_i-u(x_i)$ for $i>1$. Defining $$w(x)=u(x)+v(x-u(x))$$ we then have $w(x_1)=x_1+v(x_1-x_1)=x_1$ and $w(x_i)=u(x_i)+v(x_i-u(x_i))=x_i$ for $i>1$, so $w(x_i)=x_i$ for all $i\geq 1$.

As requested above, I am posting my comment as an answer. The implication $(\delta) \implies (\alpha)$ is addressed (modulo the equivalence of the given hypotheses with the condition that $F/K$ be a flat $R$-module) in Lemma 2.10(c) of Milne's "Etale Cohomology". A link to the relevant portion (kindly provided by Darij Grinberg) can be found here.