You can indeed make use of the proof from Hatcher Section 2.B. Presumably what that result gives you is trivial reduced homology in all dimensions for $S^n - h(D^k)$.
The way you use it is not to decompose $S^3 - (\text{Im}(f_1) \cup \text{Im}(f_2))$ but instead to decompose $S^3$ itself.
Let $U_1 = S^3 - \text{Im}(f_1)$ and let $U_2 = S^3 - \text{Im}(f_2)$, and so $U_1 \cap U_2 = S^3 - (\text{Im}(f_1) \cup \text{Im}(f_2))$.
So you want to compute the homology of $U_1 \cap U_2$. You can do this using the Mayer-Vietoris sequence for the union $S^3 = U_1 \cup U_2$.
Consider for example this portion of the sequence:
$$\underbrace{H_3(U_1)}_{\approx 0} \oplus \underbrace{H_3(U_3)}_{\approx 0} \to \underbrace{H_3(U_1 \cup U_2)}_{H_3(S^3)\approx\mathbb Z} \mapsto H_2(U_1 \cap U_2) \mapsto \underbrace{H_2(U_1)}_{\approx 0} \oplus \underbrace{H_2(U_2)}_{\approx 0}
$$
It follows that $H_2(U_1 \cap U_2) \approx \mathbb Z$. Furthermore, the proof of the Mayer Vietoris sequence is sufficiently explicit that you should be able to use it to produce an explicit 2-cycle representing the generator of $H_2(U_1 \cap U_2)$. The word "explicit" must be taken with a grain of salt, of course, because you are not being given the maps $f_1$ and $f_2$ explicitly. Really what you will do is to produce a formula for that 2-cycle that is expressed in terms of $f_1$ and $f_2$.
A similar method for $H_1(U_1 \cap U_2)$ works as well, with an even simpler outcome. It also works for $H_0(U_1 \cap U_2)$ but make sure to use reduced homology to save yourself some headaches.
View $S^1$ as the unit sphere in $\mathbb{C}$ and the disc $D^2$ as the cone on $S^1$. Thus the disc has coordinates $(z,t)$ where $z\in S^1$ and $t\in[0,1]$, and $S^1\times\{0\}$ is identified to a point. Then $\mathbb{R}P^2$ is the quotient of the disc formed by the identification $(z,1)\sim(-z,1)$. I will write the coordinates in $\mathbb{R}P^2=D^2/\sim$ with angular brackets.
We have
$$U=\{\langle z,t\rangle\in\mathbb{R}P^2\mid t\leq 1/2\},\qquad V=\{\langle z,t\rangle\in\mathbb{R}P^2\mid t>0\}.$$
The contraction of $U$ is given by $F_s\langle z,t\rangle=\langle z,(1-s)t\rangle$. The inclusion $S^1\hookrightarrow U\cap V$ is given by $z\mapsto \langle z,\frac{1}{2}\rangle$ and is a homotopy equivalence. The retraction of $V$ onto $\mathbb{R}P^1\cong \{\langle z,1\rangle\in\mathbb{R}P^2\}$ is the projection $\langle z,t\rangle\mapsto\langle z,1\rangle=\langle-z,1\rangle$, which is well-defined since $\langle z,t\rangle\in V\Rightarrow t>0$. The homotopy $G_s\langle z,t\rangle=\langle z,(1-s)t+s\rangle$ is that which is required to show that the map is a deformation retraction.
We identify $\mathbb{R}P^1$ with $S^1$ by the inverse homeomorphisms
$$\langle z,1\rangle\mapsto z^2,\qquad z\mapsto \langle \sqrt z,1\rangle=\langle-\sqrt z,1\rangle.$$
Call $\alpha: V\rightarrow S^1$ the homotopy equivalence which is the composite of the previous projection followed by this identification. Then clearly the composite $S^1\hookrightarrow U\cap V\hookrightarrow V\xrightarrow{\alpha}S^1$ which appears in your diagram is the degree $2$ map
$$z\mapsto z^2.$$
Homefully you agree with this. The point of course is that we never had a chance to consider any other map to put in your square diagram: the map $f$ was what was given to use by the geometry. One thing we should note is that we have made special use of the fact that $\alpha$ is a homotopy equivalence.
If we're feeling brave we can try to see what the maps above look like in homogeneous coordinates. Take $\mathbb{R}P^2$ as a quotient of $S^2$ by the relation $(x,y,z)\simeq(-x,-y,-z)$ and write its coordinates with square brackets as $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)\in\mathbb{R}^3$ satisfy $x^2+y^2+z^2=1$.
The quotient map $\Phi:D^2\rightarrow\mathbb{R}P^2$ is given by
$$\Phi(x,y,t)=[t\cdot x,t\cdot y,\sqrt{1-t^2}],$$
and it's easily seen that this induces a homeomorphism of $D^2/\sim$ as above onto $\mathbb{R}P^2=S^2/\simeq$. The sets which appear are
$$U=\{[x,y,z]\in\mathbb{R}P^2\mid |z|\geq\sqrt{3/4}\},\qquad V=\{[x,y,z]\in\mathbb{R}P^2\mid |z|<1\}$$
(Note the change in signs of the last coordinate!). The inclusion $S^1\hookrightarrow U\cap V$ is the map $(x,y)\mapsto\Phi(x,y,1/2)=[\frac{1}{2}x,\frac{1}{2}y,\sqrt{3/4}]$. Note that this is indeed an inclusion since any coset in its image has a unique representative with positive last coordinate. The projection $V\rightarrow \mathbb{R}P^1$ is the map $[x,y,z]\mapsto[x/\sqrt{x^2+y^2},y/\sqrt{x^2+y^2},0]$ (you might like to down the homotopy $G_s$ from before using these coordinates to show again that this map is a deformation retraction). The identification $\mathbb{R}P^1\cong S^1$ is induced using stereographic projecion. In the direction we need it is
$$[x,y,0]\mapsto (x^2-y^2,2xy).$$
The composite $S^1\hookrightarrow U\cap V\hookrightarrow V\xrightarrow{\alpha} S^1$ is now
$$(x,y)\mapsto(x^2-y^2,2xy).$$
(Notice that this is exactly the same map as before: $x+iy\mapsto (x+iy)^2$). I'll leave you to check any details.
So, to answer your second question: this is why it's better for this question to consider $D^2/\sim$ - everything is much easier!
Best Answer
For convenience I will prove that the result of deleting a small open neighbourhood of $\gamma$ is a closed disk. It is not hard to see that this is equivelant.
This can be shown by combining 3 ingredients, Mayer-Vietoris (MV), and classification of surfaces (COS) (the most general vesrion, possibly with boundary, punctures not neccesarily orientable), and the classification of 1-dimensional disk bundles over $S^1$ (CO1DB) (there is two, the trivial bundle and the Moebius band). All homology groups are taken with integer coeffecients.
1.
The first step is two show that $\gamma$ is non-separating. Suppose not for a contradiction. Then we have a decomposition of $\mathbb{RP}^2$ into two connected components $A,B$ along $\gamma$. Applying MV we have a short exact sequence
$$ 0 \rightarrow \mathbb{Z} \rightarrow H_{1}(A) \oplus H_{1}(B) \rightarrow \mathbb{Z}_2 \rightarrow 0 .$$
This implies that $b_{1}(A) + b_{1}(B) =1$. WLOG lets say $b_{1}(B)=0$. By COS, the only connected surface with one $S^1$-boundary component having $b_{1}=0$ is a disk, so $B$ is a disk. But that implies that $[\gamma]$ is trivial in $\pi_{1}(\mathbb{RP}^2)$-a contradiction.
2.
Concluding that the complement is a disk. Since, $\mathbb{RP}^2 \setminus \gamma$ has one connected component, deleting a small open neighbourhood $N(\gamma)$ of it gives a connected surface with one boundary component, homeomorphic to $S^1$. I call this boundary curve $\tau$, and set $U = \mathbb{RP}^2 \setminus N(\gamma)$. Furthermore, by CO1DB $N(\gamma)$ is homeomorphic to a Moebius band (because the boundary of $N(\gamma)$ is connected the trivial bundle cannot appear). In particular, $b_{1}(N(\gamma))=1$. So applying $MV$ for the decomposition along $\tau$ gives that $b_{1}(U)=0$. By COS, $U$ is a closed disk.
*The use of COS throughout may be reduced the following Lemma.
Lemma (consequence of COS). Let $V$ be a connected surface, possibly non-orientable, with one $S^1$ boundary component and without punctures (equivelantly compact as a topological space). Suppose that $b_{1}(V)=0$, then $V$ is a closed disk.