The following is Dummit and Foote's proof of the result (Chapter 17, Section 1):
What I don't understand is how the authors conclude in the second-to-last sentence that "the projectivity of $P_{n + 1}$ implies that we may lift the map $f_n d_{n + 1}\colon P_{n + 1}\to P'_n$ to ...". The map $d_{n + 1}'$ needn't be surjective, so how can projectivity of $P_{n + 1}$ ensure such a lift?
(I am aware only of the fact any homomorphism from a projective module factors through any surjective homomorphism.)
In general, if you have an exact row, $P$ is projective and $bu=0$, as in the diagram below:
$$\require{AMScd} \begin{CD} {} && P\\ {} @VuVV \\ A @>>a> B @>>b> C \end{CD}$$
then there exists an arrow $P\longrightarrow A$ making the diagram commute. Indeed, since $bu=0$, the image of $u$ is contained in the kernel of $b$ which is the image of $a$, and hence you can replace the diagram with the usual lifting diagram defining projectives:
$$\require{AMScd} \begin{CD} {} && P\\ {} @VuVV \\ A @>>a> \mathrm{im}\ a @>>> 0 \end{CD}$$