I will address the case where $P$ and $P'$ are acyclic (as it was noted in a comment that the general case is not true). It actually follows that $\phi$ is null-homotopic from a standard result about projective resolutions (Theorem 2.2.6 in Weibel's book, for example). Since $P$ is acyclic, it is a resolution of the zero module, in the sense that
$$
\ldots \to P_2 \to P_1 \to P_0 \to 0
$$
is exact (and similarly for $P'$). Now this standard result says that if I have projective resolutions of two modules (both the zero module in this case) and a map between those two modules (the zero map!)
$$
\require{AMScd}
\begin{CD}
\ldots @>>> P_2 @>>> P_1 @>>> P_0 @>>> 0 \\
@. @. @. @. @VVV \\
\ldots @>>> P_2 @>>> P_1 @>>> P_0 @>>> 0
\end{CD}
$$
then the vertical map lifts to a chain map $P \to P'$ so that the diagram
$$
\require{AMScd}
\begin{CD}
\ldots @>>> P_2 @>>> P_1 @>>> P_0 @>>> 0 \\
@. @VVV @VVV @VVV @VVV \\
\ldots @>>> P_2 @>>> P_1 @>>> P_0 @>>> 0
\end{CD}
$$
commutes. Furthermore, the theorem asserts that this chain map is unique up to chain homotopy, meaning that any two chain maps $P \to P'$ making the diagram above commute must be chain homotopic. Well, your chain map $\phi$ is one such chain map, but so is the zero chain map. Thus $\phi$ and the zero chain map are homotopic.
Let me also indicate how to go about proving the result directly. It's similar to how you prove the theorem I cited, and in general it's the way you approach a statement about bounded below complexes of projectives: by building your maps inductively. First, take your $h_{-1} : 0 \to P_0'$ to be the zero map (your only option). Now we're looking for a map $h_0 : P_0 \to P_1'$ such that $d_1' h_0 = \phi_0$; since $d_1'$ is surjective ($P'$ being acyclic), we immediately get such an $h_0$ from the lifting property of the projective module $P_0$.
For the inductive step, we're looking for $h_n : P_n \to P_{n+1}'$ such that $d_{n+1}' h_n = \phi_n - h_{n-1} d_n$. Now, $\phi_n - h_{n-1} d_n$ lands in $\ker(d_n')$ (check this!), which is the image of $d_{n+1}'$. From here, can you see how to produce $h_n$?
I have seen that it is actually possible to construct this map if you follow the proof in Gelfand-Manin carefully. In there, they say that the way to construct the map between Cartan-Eilenberg resolutions is to apply Theorem III.1.3 and statement III.7.11.B (inside the proof) repeatedly.
- Theorem III.1.3 asserts that it is possible to extend any map between objects to a map between projective resolutions of said objects. Inverting the arrows gives the statement we need for injective resolutions.
- Remark III.1.4(b) points out that if we aim to extend $f: X \to Y$, the resolution of $Y$ need not actually be projective, just an exact sequence. Dually, we don't need the resolution on the source to be injective.
This is precisely our setting: $T$ being exact guarantees that all exactness properties of $I^{\bullet, \bullet}$ pass on to $T(I^{\bullet, \bullet})$, which is all we needed to carry on with the construction.
Best Answer
I can only sketch a solution. First consider what $\tilde{f}$ could be if $f=0$ (not homotopic to zero but just zero). Then what would happen is that the C-E (Cartan Eilenberg) resolution over H and B (the homology and boundary projective resolutions which generate the C-E resolution) would be a chain map over zero. What are the possible chain maps over projective resolutions over zero? They must be chain homotopic to zero. There's a version of horseshoe lemma which allows you to extend these homotopies to the C-E resolution, this gives a vertical homotopy to zero over any identically zero map. Now returning to the question, we wonder what happens if we are a C-E resolution over a homotopy zero map. In this case, simply lift the horizontal homotopy to projective resolutions to get a C-E map which is horizontally homotopic to zero. Now subtract this C-E map from the one given, it must be a C-E map over zero which is vertically homotopic to zero. This shows that any C-E map over a chain map homotopic to zero must be homotopic to zero.