I think the answer is:
$$ H_3 (T_f) = 0, \qquad H_2 (T_f) = \mathbb Z_2, \qquad H_1 (T_f) = \mathbb Z, \qquad H_0 (T_f) = \mathbb Z.
$$
The only thing we have to care about is finding the effect of $\ f_*\ $ on the homology groups $\quad H_2 (S^2) = \mathbb Z \quad $ and $\quad H_0 (S^2) = \mathbb Z. \quad $
Now the $0$-th induced morphism is just the identity: of course $f$ maps the sphere to itself, and we know that the $0$-th homology group is just the free $\mathbb Z$-module generated by the connected components of our space (cfr. E.H. Spanier "$Algebraic \ Topology$" Chapter 4, pag 155).
So, in our exact sequence
$$
H_1 (S^2)\longrightarrow H_1 (T_f) \longrightarrow H_0 (S^2)\xrightarrow{ 1 - f_* } H_0 (S^2) \longrightarrow H_0 (T_f) \longrightarrow 0
$$
we can substitute
$$ H_1 (S^2) = 0, \qquad H_0 (S^2) = \mathbb Z, \qquad 1-f_* = 0, \qquad $$
to obtain
$$
0 \longrightarrow H_1 (T_f) \longrightarrow \mathbb Z \xrightarrow{\ \ 0 \ \ } \mathbb Z \longrightarrow H_0 (T_f) \longrightarrow 0.
$$
Hence,
$$
H_1 (T_f) = \mathbb Z, \qquad H_0 (T_f) = \mathbb Z.
$$
It remains to see
$$
... \longrightarrow H_3 (S^2)\longrightarrow H_3 (T_f) \longrightarrow H_2 (S^2)\xrightarrow{ 1 - f_* } H_2 (S^2) \longrightarrow H_2 (T_f) \longrightarrow H_1 (S^2) \longrightarrow ...
$$
Of course
$$
H_3 (S^2) = H_1 (S^2) = 0.
$$
We need to know what is $$f_*: H_2 (S^2)\rightarrow H_2 (S^2). $$
Now $\quad 1 \in H_2 (S^2) \quad $ can be tought of as the orientation class of $S^2$ (see, ad example https://en.wikipedia.org/wiki/Orientability#Homology_and_the_orientability_of_general_manifolds). For $n$ even the antipodal map is orientation-reversing (M. P. Do Carmo $Riemannian \ Manifolds$, Chpt 0, page 20), and, since it is a cover of order one, it has to induce the map $-1$ in homology.
This is also directly stated in E.H. Spanier "$Algebraic \ Topology$" (Chapter 4, pag 196, section 7, points 9-10), which should be a credible enough reference.
So, since $\ 1 - (-1) = 2\ $ (just kidding), and $\ H_2 (S^2) = \mathbb Z $
$$
0 \longrightarrow H_3 (T_f) \longrightarrow \mathbb Z\xrightarrow{\quad 2 \times \quad } \mathbb Z \longrightarrow H_2 (T_f) \longrightarrow 0
$$
we have $H_3 (T_f) = \ker (2 \times ) = 0 \ $ and $\ H_2 (T_f) = \mathbb Z / 2 \mathbb Z = \mathbb Z_2.$
A couple comments. The result is quite clear: $f$ is orientation-reversing, so one should expect $T_f$ to be a non-orientable manifold, i. e. $H_3 (T_f) = 0. \ $ $H_0$ and $H_1$ are obvious, since the mapping torus is a $S^2$-fibration over $S^1$. As for $H_2$, you see that two copies of $S^2$ will cancel each other in $T_f$: just think one as the inverted copy of the other.
Second thing: if you want to learn algebraic topology, you really have to study
on the textbook by E. H. Spanier. It's old but gold.
I'm not 100% certain but I'm pretty sure this is the idea of how to solve it without breaking up the sequence as you've suggested.
Using your statement about $H_k(S^1 \times S^1)$, and your first long exact sequence, because $H_k(S^1 \times S^1) = 0$ for $k \geq 3$, the nontrivial part of the exact sequence becomes
$$0 \to H_3(T_f) \to \mathbb Z \xrightarrow[]{\mathbb 1 - f_*} \mathbb Z \to H_2(T_f) \to \mathbb Z^2 \xrightarrow[]{\mathbb 1 - f_*} \mathbb Z^2 \to H_1(T_f) \to \mathbb Z \xrightarrow{\mathbb1 - f_*} \mathbb Z \to H_0(T_f) \to 0,$$
where $\mathbb 1$ is the identity map. Since $T_f$ has one connected component, $H_0(T_f) = \mathbb Z$. The dimension-$2$ induced homology map is $1 \mapsto -1$. (One way of seeing this is by giving $S^1 \times S^1$ a simplicial structure as in Example 2.3 in Hatcher, and observing that $f_\sharp(U - L)$ is homologous to $-U + L$.) Therefore the left hand map $\mathbb 1 - f_* : \mathbb Z \to \mathbb Z$ is defined by $1 \mapsto 2$. Since this is injective, exactness gives us $\tilde H_3(T_f) = 0$.
The dimension-$0$ induced homology map $f_* : \mathbb Z \to \mathbb Z$ is the identity, so $\mathbb 1 - f_* = 0$ in the right hand map $\mathbb Z \to \mathbb Z$. Meanwhile, the dimension-$1$ induced homology map $f_* : \mathbb Z^2 \to \mathbb Z^2$ sends one generator to itself, and the other generator to its inverse, and so can be represented by the matrix $\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$. Therefore $\mathbb 1 - f_* = \left( \begin{array}{cc} 0 & 0 \\ 0 & 2 \end{array}\right)$. By exactness, $\ker(\mathbb Z^2 \to H_1(T_f)) = \mathrm{im}\left( \begin{array}{cc} 0 & 0 \\ 0 & 2 \end{array}\right)$, which is generated by $2$, so $\mathbb Z^2 / \ker(\mathbb Z^2 \to H_1(T_f)) = \mathbb Z \oplus \mathbb Z_2$. This gives us a short exact sequence
$$0 \to \mathbb Z \oplus \mathbb Z_2 \to H_1(T_f) \to \mathbb Z \to 0.$$
Since $\mathbb Z$ is free, this exact sequence splits, meaning $H_1(T_f)$ is the direct sum of the first and third nontrivial terms. So we have $H_1(T_f) = \mathbb Z^2 \oplus \mathbb Z_2$.
Lastly, recall $\mathbb Z \to \mathbb Z$ on the left side of the first exact sequence is $1 \mapsto 2$. By exactness, this is also the kernel of $\mathbb Z \to H_2(T_f)$, so by the first isomorphism theorem,
$$ \mathrm{im}(\mathbb Z \to H_2(T_f)) = \mathbb Z / 2\mathbb Z = \mathbb Z_2.$$
Also, the map $H_2(T_f) \to \mathbb Z^2$ has image generated by $(1,0)$, and again by the first isomorphism theorem and exactness, $$\mathbb Z = \mathrm{im}(H_2(T_f) \to \mathbb Z^2) = H_2(T_f)/\ker(H_2(T_f) \to \mathbb Z^2) = H_2(T_f) / \mathrm{im}(\mathbb Z \to H_2(T_f)) = H_2(T_f) / \mathbb Z_2.$$
By uniqueness of the decomposition of a finitely generated abelian group into cyclic subgroups, this gives us $H_2(T_f) = \mathbb Z \oplus \mathbb Z_2$. So, to summarize,
$$ H_n(T_f) = \begin{cases}
\mathbb Z & \textrm{if } n = 0, \\
\mathbb Z^2 \oplus \mathbb Z_2 & \textrm{if } n = 1, \\
\mathbb Z \oplus \mathbb Z_2 & \textrm{if } n = 2, \\
0 & \textrm{if } n \geq 3.
\end{cases}
$$
Best Answer
Consider the exact sequence \begin{align}\ldots\longrightarrow H_{k}(\Bbb RP^\infty;\Bbb Z)\longrightarrow H_{k}(\Bbb RP^\infty;\Bbb Z/2)\overset{\bar{\beta_2}}{\longrightarrow}H_{k-1}(\Bbb RP^\infty;\Bbb Z)\overset{f}{\longrightarrow}H_{k-1}(\Bbb RP^\infty;\Bbb Z)\\ \overset{r_{\ast}}{\longrightarrow} H_{k-1}(\Bbb RP^\infty;\Bbb Z/2)\longrightarrow \ldots\end{align} Suppose $k\ge 2$ first.
If $k$ is even, then $H_{k}(\Bbb RP^\infty;\Bbb Z)\cong 0$ so that $\bar{\beta_2}$ is injective and hence an isomorphism because $H_{k}(\Bbb RP^\infty;\Bbb Z/2)$ and $H_{k-1}(\Bbb RP^\infty;\Bbb Z)$ are both isomorphic to $\Bbb Z/2$, which forces $f$ to be trivial so that $r_*$ is also an isomorphism. Therefore, $\beta_2=r_*\bar{\beta_2}$ is also an isomorphism.
If $k$ is odd, then $H_{k-1}(\Bbb RP^\infty;\Bbb Z)\cong 0$, which means that $\bar{\beta_2}$ trivial so that $\beta_2=r_*\bar{\beta_2}$ must also be trivial.
Suppose $k=1$, then we are computing the map $\beta_2:H_1(\Bbb RP^\infty;\Bbb Z/2)\to H_0(\Bbb RP^\infty;\Bbb Z/2)$. The long exact sequence for this part looks like the following. \begin{align}\ldots\longrightarrow H_1(\Bbb RP^\infty;\Bbb Z/2)\overset{\bar{\beta_2}}{\longrightarrow}H_0(\Bbb RP^\infty;\Bbb Z)\overset{f}{\longrightarrow} H_0(\Bbb RP^\infty; \Bbb Z)\overset{r_*}{\longrightarrow}H_0(\Bbb RP^\infty;\Bbb Z/2)\longrightarrow 0\end{align} Since we have $H_1(\Bbb RP^\infty;\Bbb Z/2)\cong\Bbb Z/2$ and $H_0(\Bbb RP^\infty;\Bbb Z)\cong \Bbb Z$, the map $\bar{\beta_2}$ has to be trivial, and $\beta_2=r_*\bar{\beta_2}$ must also be trivial.
In conclusion, $\beta_2:H_k(\Bbb RP^\infty;\Bbb Z/2)\to H_{k-1}(\Bbb RP^\infty;\Bbb Z/2)$ is an isomorphism for $k$ even and trivial for $k$ odd.