In page 48 of Charles Weibel's book 'an introduction to homological algebra.'https://people.math.rochester.edu/faculty/doug/otherpapers/weibel-hom.pdf I find the following statement but I do not fully understand the proof.
Theorem Assume $A$ has enough projectives. Then for any right exact functor $F$, the derived functor $LF$ forms a universal $\delta$-functor.
Weibel uses in his book induction to prove this statement. Assume $T$ is another $\delta$-functor and the natural transformation which commutes with $\delta$ have been constructed for $0\leq i\leq n-1$. Then I think in the induction step we need to prove for any short exact sequence $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$
there exist three vertical maps on the left such that the diagram commutes.
$$\require{AMScd}
\begin{CD}
T_{n}(A') @>>> T_n(A) @>>> T_n(A'') @>>> T_{n-1}(A') @>>> T_{n-1}(A) @>>> T_{n-1}(A'') \\
@Vf_n(A')VV @Vf_n(A)VV @Vf_n(A'')VV @Vf_{n-1}(A')VV @Vf_{n-1}(A)VV @Vf_{n-1}(A'')VV \\
L_{n}F(A') @>>> L_nF(A) @>>> L_nF(A'') @>>> L_{n-1}F(A') @>>> L_{n-1}F(A) @>>> L_{n-1}F(A'')\\
\end{CD}
$$
But what Weibel did is just to prove there exists $f_{n}(A'')$ such that three squares in the following diagram (or three squares on right of the diagram above) commutes. Did I make a mistake or did I understand incorrectly? $$\require{AMScd} \begin{CD} T_n(A'') @>>> T_{n-1}(A') @>>> T_{n-1}(A) @>>> T_{n-1}(A'') \\ @Vf_n(A'')VV @Vf_{n-1}(A')VV @Vf_{n-1}(A)VV @Vf_{n-1}(A'')VV \\ L_nF(A'') @>>> L_{n-1}F(A') @>>> L_{n-1}F(A) @>>> L_{n-1}F(A'')\\ \end{CD} $$
What is required is to construct a natural transformation $\varphi_n: T_n \to L_n$ such that the collection $\{\varphi_n\}$ are a morphism of $\delta$-functors. To construct $\varphi_n$ we need for each $A$, a map $\varphi_n(A): T_n(A) \to L_n(A)$ as the components of the natural transformation.
Weibel's proof takes any $A$ and places it into an exact sequence as $0 \to K \to P \to A \to 0$. Then, using this exact sequence, he constructs a unique map $\varphi_n(A): T_n(A) \to L_n(A)$ compatible with the prior $\varphi_m$ for $m < n$.
Next, his proof shows that the construction of this map $\varphi_n$ does not depend on the choice of $P$, and that it is functorial in $A$ (which is done by considering $f: A \to A'$ and showing that $\varphi_n$ commutes with $f$). This gives the commutativity of left two squares of the first diagram that you display.
Finally, one must check that $\varphi_n$ is compatible with the prior choice of $\varphi_{n-1}$, so that it can give a map of $\delta$-functors. This is commutativity of the leftmost square of the second diagram you display, which together with the commutativity of the left two squares of the first diagram (as was already shown) and commutativity of the right two squares (by the inductive hypothesis) gives commutativity of the first diagram.