Apply the snake lemma not to
$$\begin{array}{ccccccccc}
& & A_q & \rightarrow & B_q & \rightarrow & C_q & \rightarrow & 0 \\
& & \downarrow & & \downarrow & & \downarrow & & \\
0 & \rightarrow & A_{q-1} & \rightarrow & B_{q-1} & \rightarrow & C_{q-1}, & &
\end{array}$$
but to
$$\DeclareMathOperator{\coker}{coker}
\begin{array}{ccccccccc}
& & \coker(\alpha_{q+1}) & \rightarrow & \coker(\beta_{q+1}) & \rightarrow & \coker(\gamma_{q+1}) & \rightarrow & 0 \\
& & \downarrow & & \downarrow & & \downarrow & & \\
0 & \rightarrow & \ker(\alpha_{q-1}) & \rightarrow & \ker(\beta_{q-1}) & \rightarrow & \ker(\gamma_{q-1}). & &
\end{array}$$
Make sure you see where the maps come from, that the diagram commutes, and that the rows are exact.
Here's the diagram you wanted to include:
This represents how the connecting homomorphism $d^{*} \colon H^{k}(\mathcal{C}) \to H^{k+1}(\mathcal{A})$ is defined: you take a representative cocycle $c$ of an element $[c] \in H^{k}(\mathcal{C}),$ you find an element $b \in B^{k}$ such that $j(b) = c$ (this is possible because $j$ is surjective, by exactness). Then, you show that $j(d(b)) = 0,$ hence there exists $a \in A^{k+1}$ such that $i(a) = d(b)$ (by exactness, $\ker j = \text{im} \; i$). Finally, you check that $d(a) = 0,$ hence $a$ is a cocycle, and $[a]$ is an element of $H^{k+1}(\mathcal{A}).$ You define
$$d^{*}([c]) = [a].$$
The details of the above definition can be found in Section 25.3 of Tu. A very important exercise is 25.5, which asks you to show that the above definition is well-defined. Indeed, we've made a number of choices: there are possibly multiple $b, b'$ such that $j(b) = j(b') = c,$ and there are possibly multiple $c, c'$ such that $[c] = [c'].$ It is very important that you check for yourself that these choices end up giving the same $[a]$ at the end. If you need any hints, I can include some in this answer.
Going back to your question, the well-definedness of the above process is key. Since the result of the function $d^{*}$ is going to be the same no matter which representative $c$ we choose and which element $b$ we choose, we might as well pick convenient ones.
Consider $[b] \in H^{k}(\mathcal{B}).$ Then,
$$d^{*}(j^{*}([b])) = d^{*}([j(b)]).$$ Remember that to define $d^{*}([c]),$ we first pick a representative for $[c].$ Here, our $[c]$ is $[j(b)],$ so we might as well pick the representative $j(b)$. Since $d^{*}$ is well-defined, we can make this choice to be the most convenient for us.
Now, following the definition of $d^{*}$, we want to find some element of $B^{k}$ such that $j$ sends that element to our representative $j(b).$ We might as well pick our element to be $b,$ because $j$ sends $b$ to $j(b).$ Once again, since $d^{*}$ is well-defined, we can make this choice to be the most convenient for us.
Then, we send $b$ to $d(b),$ and we find an $a \in A^{k+1}$ such that $i(a) = d(b).$ Note that $b$ is a cocycle, so $d(b) = 0.$ So, we want to find $a \in A^{k+1}$ such that $i(a) = 0.$ Since $i$ is a homomorphism, $a = 0$ does the trick (in fact, since $i$ is injective, this is the only possible $a$).
Therefore,
$$d^{*}([j(b)]) = [a] = [0] = 0.$$ The conclusion follows.
Best Answer
So suppose that $c$ and $c'$ are two alternative representatives of the class $[c]$. We follow your construction through for both $c$ and $c'$. Concretely, this means we find $b$ and $b'$ such that $j(b) = c$ and $j(b') = c'$, and we find $a$ and $a'$ such that $i(a) = d(b) $and $i(a') = d(b')$.
Our task is to show that $a$ and $a'$ represent the same class in $H^{k+1}(\mathcal A)$. In other words, our task is to find a $w \in A^k$ such that $d(w) = a' - a$.
Let's construct this $w$.
First, the fact that $c$ and $c'$ represent the same class in $H^k (\mathcal C)$ means that there exists a $u \in C^{k-1}$ such that $d(u) = c' - c$.
Since $j : B^{k-1} \to C^{k-1}$ is surjective, there exists a $v \in B^{k-1}$ such that $j(v) = u$.
Now consider the element $z := b' - b - d(v)$. Notice that $$ j(z) = j(b') - j(b) - j(d(v)) = j(b') - j(b) - d(j(v)) = j(b') - j(b) - d(u) \\ = c' - c - (c' - c) = 0 $$
So $z$ is in the kernel of $j$. By exactness, $z$ must also be in the image of $i$, i.e. there must exist a $w \in A^k$ such that $i(w) = z$.
Now observe that $$ i(d(w) - (a' - a)) = d(i(w)) - (i(a') - i(a)) = d(z) - (i(a') - i(a)) \\ = (d(b') - d(b) - d(d(v))) - (d(b') - d(b)) = 0. $$
Since $i$ is injective, we conclude that $$ d(w) = a' - a.$$ Thus we've constructed the $w$ that we set out to construct! This $w$ is evidence to the fact that $[a] = [a']$ in $H^{k+1}(\mathcal A)$.