To make sense of a proof of this nature, it isn't enough to know that there exists a map from $H_d(V, V \backslash \{ y \})$ to $H_d(S^d, S^d \backslash \{ y \})$, and that there exists an isomorphism between $H_d(S^d)$ and $H_d(S^d, S^d \backslash \{ y \})$, and so on. You really also need understand how all of these maps act on the cycles in the various homology groups.
Sometimes (but not always), it is possible to describe how a map acts on cycles in homology groups by identifying the map on homology as the map induced by a continuous function between the relevant topological spaces. Since we usually have a good idea of how to compose continuous functions between topological spaces, this will often enable us to work out how to compose the maps on the homology groups. In certain situations, this will also enable us to prove that certain maps between homology groups are identity maps, or zero maps.
As it happens, all of the maps in your diagram are induced by continuous functions between topological spaces:
The horizontal maps are all induced by $f$ (or restrictions of $f$ to the appropriate spaces).
The top-left vertical map is induced by the identity map $S^d \to S^d$. The same applies to the top-right vertical map. This is just how these maps are defined.
The bottom-left vertical map is induced by the various identity maps $U_i \to U_i$.
The middle-left vertical map is induced by the various inclusion maps $U_i \to S^d$. The bottom-right vertical map is induced by the inclusion map $V \to S^d$. The fact that these maps are induced by the relevant inclusion maps is a part of the statement of the excision theorem, and it is worth noting this!
It's clear that the diagram in your book really is a commutative diagram, because the corresponding maps between topological spaces all commute appropriately!
Each $H_d(U_i, U_i \backslash \{ x_i \} ) $ is isomorphic to $\mathbb Z$, since $$H_d(U_i, U_i \backslash \{ x_i \} ) \cong H_d(S^d, S^d \backslash \{ x_i \}) \cong H_d (S^d) \cong \mathbb Z,$$
where the first equality is by excision and the second equality is by the LES for the pair $(S^d, S^d \backslash \{ x_i \})$.
Therefore, $$H_d (S^d, S^d \backslash \{ x_1, \dots, x_k \} ) \cong \oplus_i H_d(U_i , U_i \backslash \{ x_i \}) \cong \mathbb Z^{\oplus k}.$$
Now let us define the cycle $$(0, \dots, 1, \dots, 0) \in H_d (S^d, S^d \backslash \{ x_1, \dots, x_k \} )$$
(with the $1$ in the $i$th position) to be generator coming from the generator $1 \in H_d(U_i, U_i \backslash \{ x_i \} ) \cong \mathbb Z$.
Here's an important question we must address, if we're to make any progress:
Given a cycle $$(a_1, \dots, a_k) \in H_d (S^d, S^d \backslash \{ x_1, \dots, x_k \} ),$$ is there a simple way to determine the numbers $a_1, \dots, a_k$, if we don't already know them?
And here's my proposed solution:
For each $i \in \{ 1, \dots, k \}$, define a natural map $$p_i : H_d(S^d, S^d \backslash \{ x_1, x_2, \dots, x_k \}) \to H_d(S^d, S^d \backslash \{ x_i \}), $$ to be the map on homology induced by the identity map $S^d \to S^d$.
Then for each $i$, $$ p_i(a_1, \dots, a_k) = a_i \in H_d(S^d, S^d \backslash \{ x_i \}).$$
Let's prove this carefully. First, take $$(1,0,\dots , 0) \in H_d(S^d, S^d \backslash \{ x_1, \dots, x_k \} )$$ and map it to $H_d(S^d, S^d \backslash \{ x_1 \})$ via $p_1$.
Since $(1,0,\dots, 0)$ is itself the image of the generator $1 \in H_d(U_1, U_1 \backslash \{ x_1 \})$, we know that $p_1(1,0,\dots, 0)$ is the image of $1 \in H_d(U_1, U_1 \backslash \{ x_1 \})$ under the composition,
$$ H_d(U_1,U_1 \backslash \{ x_1 \}) \to H_d(S^d, S^d \backslash \{ x_1, \dots, x_k \} ) \to H_d(S^d, S^d \backslash \{ x_1 \}).$$
The first map is induced by the inclusion $U_1 \to S^d$ and the second map is induced by the identity map $S^d \to S^d$, so the composition is induced by the inclusion $U_1 \to S^d$.
But we know that the map $H_d(U_1,U_1 \backslash \{ x_1 \}) \to H_d(S^d, S^d \backslash \{ x_1 \})$ induced by the inclusion $U_1 \to S^d$ is an isomorphism, by excision! So we conclude that $$p_1(1,0, \dots, 0) = 1 \in H_d(S^d, S^d \backslash \{ x_1 \}).$$
Okay, how about we take $$(1,0,\dots , 0) \in H_d(S^d, S^d \backslash \{ x_1, \dots, x_k \} )$$ as before, but this time, we map it via $p_2$ to $H_d(S^d, S^d \backslash \{ x_2 \})$?
The image $p_2(1,0 \dots, 0)$ is the same as the image of $1 \in H_d(U_1, U_1 \backslash \{ x_1 \})$ under the composition,
$$ H_d(U_1,U_1 \backslash \{ x_1 \}) \to H_d(S^d, S^d \backslash \{ x_1, \dots, x_k \} ) \to H_d(S^d, S^d \backslash \{ x_2 \}).$$
But if you think about it, this composition is the same as the composition,
$$ H_d(U_1,U_1 \backslash \{ x_1 \}) \to H_d(U_1,U_1) \to H_d(S^d, S^d \backslash \{ x_2 \}),$$
where the first map is induced by the identity $U_1 \to U_1$ and the second map is induced by the inclusion $U_1 \to S^d$. (Note that since $U_1 \subset S^d \backslash \{ x_2 \}$, the second map really is well-defined.) And why are these two compositions equal? Because both of these compositions are the maps on homology induced by the inclusion map $U_1 \to S^d$!
Of course, $H_d(U_1, U_1) = 0$, so it clear that $$p_2(1,0,\dots 0 ) = 0 \in H_d(S^d, S^d \backslash \{ x_2 \}).$$ This completes the proof of my claim.
Right. Having done all this hard work, we're going to prove that the image of $1 \in H_d(S^d)$ under the top-left map in the diagram is the element $$(1,1, \dots 1) \in H_d(S^d, S^d \backslash \{ x_1, \dots, x_k \} ). $$
I believe this is the part of the proof that you weren't sure about.
By the claim that we just proved, we only have to verify that the image of $1 \in H_d(S^d)$ under the composition
$$ H_d (S^d) \to H_d(S^d, S^d \backslash \{ x_1, \dots, x_k \} ) \overset{p_i} {\to} H_d(S^d, S^d \backslash \{ x_i \})$$
is the element $$1 \in H_d(S^d, S^d \backslash \{ x_i \}).$$
This is straightforward to show. The composition I wrote down is the map on homology induced by the identity map $S^d \to S^d$. But the map $H_d (S^d) \to H_d(S^d, S^d \backslash \{ x_i \})$ induced by the identity map $S^d \to S^d$ is precisely the map appearing in the LES for the pair $(S^d, S^d \backslash \{ x_i \})$, and this map is an isomorphism.
So $1 \in H_d(S^d)$ maps to $1 \in H_d(S^d, S^d \backslash \{ x_i \})$, and we're done.
To finish off, ${\rm deg}f$ is the image of $1 \in H_d(S^d)$ under the map $f_\star$. By the commutative diagram, this is the same as the image of $(1,1,\dots, 1) \in H_d(S^d, S^d \backslash \{ x_1, \dots, x_k \})$ under $f_\star$. And this is the same as the sums of the images of $1 \in H_d( U_i, U_i \backslash \{ x_i \})$ under $f_\star$.
Thus we have shown that
$${\rm deg} f = \sum_i {\rm deg} f_i.$$
Best Answer
This is the answer proposed by Steve D.
By the homotopy axiom, the map $H(f):H_1(S^1) \rightarrow H_1(S^1)$ is the same for homotopic maps $g \simeq f$.
When $|a|<1$, we give the homotopy $$ H_t(z):= \frac{z-ta}{|z-ta|}.$$ So the degree of $f$ is equivalent to that of $g(z)=z$, which is $1$.
When $|a|>1$, we give the homotopy $$H_t(z):= \frac{tz-a}{|tz-a|}$$ so the degree of $f$ is equivalent to that of the constant map, which has degree $0$.