Let $0\rightarrow M'\rightarrow M \rightarrow M''\rightarrow 0$ be a short-exact sequence of modules (over a ring $R$). By a projective resolution of this sequence, we mean (according to Jacobson volume II, p. 344):
(i) A projective resolution $\cdots \rightarrow C_1'\rightarrow C_0'\rightarrow M'\rightarrow 0$ of $M'$;
(ii) A projective resolution $\cdots \rightarrow C_1\rightarrow C_0\rightarrow M\rightarrow 0$ of $M$;
(iii) A projective resolution $\cdots \rightarrow C_1''\rightarrow C_0''\rightarrow M''\rightarrow 0$ of $M''$;
(iv) maps $i_n:C_n'\rightarrow C_n$ and $p_n:C_n\rightarrow C_n''$
such that
(I) for every $i\ge 0$, $0\rightarrow C_i'\rightarrow C_i\rightarrow C_i''\rightarrow 0$ is a short exact sequence.
(II) The maps between $C_0',C_0,C_0''$ and $M',M,M''$ [which appear in (i)-(iv)] make two squares commutative.
Q. Is it not necessary in the definition that maps between $C_n',C_n,C_n''$ and $C_{n-1}',C_{n-1},C_{n-1}''$ [which appear in (i)-(iv)] make two squares commutative?
To look at question in simple way, think of three sequences in (i)-(iii) are written exactly one below the other; then it is given that the column of $M$'s is exact. Then in the definition of projective resolution of this exact sequence of $M$'s, we want
the column of $C_0$'s, column of $C_1$'s, etc. exact,
the column of $C_0$'s and $M$'s makes two squares commutative.
My question is, don't we require (in definition) the column of each $C_n$'s and $C_{n-1}$'s make two commutative squares?
As far the argument in Jacobson's book is considered, the author constructs the projective resolution of $M$ in such a way that the maps in that resolution (of $M$) make all small squares commutative.
As requested, this answer is the same as a previous comment.
It is in fact stated that a resolution of $0\to M′\to M\to M''\to 0$ is the data of resolutions $C′,C,C′′$ of $M′,M,M′′$ together with chain homomorphisms $C′\to C,C\to C′′$ such that the conditions (I) and (II) in the OP hold. The fact that these are chain homomorphisms means that every square must commute.