On the top of page 789 (I own the hardcopy of the book btw ;) it says:
Next we must construct $f_1$. We write the second square in the form $$ \require{AMScd} \begin{CD} 0 @>>> E^0/M @>>> E^1\\ & @V{f_0}VV\\ & & I^0 @>>> I^1 \end{CD} $$
Now on the previous page we have proven (and understood the proof) that:
$$ \require{AMScd} \begin{CD} 0 @>>> M @>>> E^0 @>>> E^1 @>>> \dots \\ & & @V{\varphi}VV @V{f_0}VV\\ 0 @>>> M' @>>> I^0 @>>> I^1 @>>> \dots \end{CD} $$
commutes, i.e. that such a morphism $f_0$ that makes the first square commute. The overall lemma I'm trying to prove is given the above diagram (minus $f_0$), where the top row is exact and the bottom row is such that each $I^n$ is an injective object, then there exists a morphism of the two complexes $f: E \to I$ such that $f_{-1} = \varphi$ and for any two such morphisms $f,g$ of complexes, they are homotopic to one another, meaning there exists morphisms (in the ambient abelian category) $h_n : E^n \to I^{n-1}$ such that $f_n - g_n = d'^nh_{n+1} + h_{n + 2}d^{n+1}$ where $d$ is the differential for the top row (which is exact), and $d'$ the differential for the bottom row.
So what does Lang mean by "write the second square in the form ..."?
The top row is exact so that $d^{-1} : M \rightarrowtail E_0$ is an injection. By $E_0 / M$ he actually means $E_0 / d^{-1}(M) = E_0 /\ker d_0$.
Because we quotiented by the kernel of $d_0$, we get that the induced map ${d^0}^*: E_0/\ker d^0 \rightarrowtail E_1$ is an injection, so that we get:
$$ \require{AMScd} \begin{CD} E_0 @>{\pi}>> E_0/\ker d^0 @>{d^0}^*>> E^1\\ @V{f_0}VV @V{f_0^*}VV @V{\exists f_1}VV\\ I^0 @>{=}>> I^0 @>{d'^0}>> I^1 \end{CD} $$
Now you compose $g = d'^0 \circ f_0^*$ to get an injectivity triangle on the right, since ${d^0}^*$ is injective (a monic). By definition of injectivity, there exists $f_1 : E^1 \to I^1$ making the triangle/square commute. The square on the left also commutes by definition of induced map $f_0^*$ on the quotient. When you paste together two commuting polygons along a common edge, the resulting diagram commutes. Thus composing all the maps, we get that the outer rectangle commutes, which is the commuting square we're after, the second one in the sequence.