Consider a left exact functor $F: \mathcal{A}\to \mathcal{B}$. According to the Wikipedia article https://en.wikipedia.org/wiki/Derived_functor the right derived functors $R^iF$ of $F$ are defined as follows. For an object $A$ in $\mathcal{A}$, one considers an injective resolution $0\to A\to I_0 \to I_1 \to I_2\cdots .$ Applying the functor $F$ to the above and "chopping off $A$" gives $0\to F(I_0)\to F(I_1) \to F(I_2)\to \cdots.$ Then $R^iF(A)$ is the $i$-th cohomology group of the above complex. Now if $R^iF$ is a functor, for every morphism $\phi:A\to A'$ in $\mathcal{A}$ there should exist a morphism $R^iF(\phi):R^iF(A)\to R^iF(A')$. My question is how to construct the morphisms $R^iF(\phi)$?
After spending some time with it I think I have an answer. But I am not confident enough to put it as an answer in case there are some loopholes. So, I am posting it here as an edit.
In order to show that there exist maps $R^iF(\phi): R^iF(A)\to R^iF(B)$, we have to show that there exists a map between the complexes $A_\bullet\to B_\bullet$ as shown by the following commutative diagram: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} 0 &\to & A & \ra{i} & I_0 & \ra{d_0} & I_1 & \ra{d_1} & I_2 & \ra{d_2} & \cdots \\ \da{0} & & \da{\phi} & & \da{\psi_0} & & \da{\psi_1} & & \da{\psi_2} & & \da{\psi_n} \\ 0 & \to & B & \ra{i'} & I_0' & \ra{d_0'} & I_1' & \ra{d_1'} & I_2' & \ra{d_2'} & \cdots \\ \end{array} $$ Our goal is to construct the maps $\psi_n$. Construction of $\psi_0$ is easy; we basically lift the map $i'.\phi$ to a map $\psi_0: I_0 \to I_0'$ which is possible since $I_0'$ is injective. Before proceeding further let's show the following fact:
If $a_0\in image(i)\subset I_0$ then $\psi_0(a_0)\in image(i')$.
Proof: Let $a_0\in image(i)$ $\implies$ there exists an $a\in A$ such that $i(a)=a_0$. This implies that $ i'.\phi(a)=\psi_0(a_0)$. Hence $\psi_0(a_0)\in image(i')$.
This also implies that if $a_0\in ker(d_0)$ then $\psi_0(a_0)\in ker(d_0')$.
We now proceed to construct the map $\psi_1: I_1 \to I_1'$. We first construct the map $\psi_1':image(d_0)\to I_1'$ in the following way:
For any $a_1\in image(d_0)$ we choose an $a_0\in I_0$ such that $d_0(a_0)=a_1$ and set $\psi_1'(a_1)=d_0'.\psi_0(a_0)~|~ d_0(a_0)=a_1$. To show that $\psi_1'(a_1)$ is independent of the choice of $a_0 \in I_0$, let's choose the representative $a_0+k_0~(k_0\in ker(d_0))$ instead of $a_0$, so that $\psi'(a_1)=d_0'.\psi_0(a_0+k_0)=d_0'.\psi_0(a_0)+d_0'.\psi_0(k_0)=d_0'.\psi_0(a_0)$, since $\psi_0(k_0)\in ker(d_0')$ as shown above.
Having constructed the map $\psi_1': image(d_0) \to I_1'$, we can extend it to $\psi_1: I_1 \to I_1'$ since $I_1'$ is again injective. We can now construct all the maps $\psi_n$ recursively following the same procedure.
If anyone can check and let me know whether the above construction is correct or there are some loopholes in it, that will be very helpful.