Consider a map $f\colon S^1 \to S^1 \vee S^1$ that maps upper half of a circle to the first summand in orientation preserving way, and lower half to the second circle in orientation reversing way. Also denote $i_1$ and $i_2$ inclusions of $S^1$ to $S^1 \vee S^1$ as two summands. Composition of $f$ with folding map $p\colon S^1\vee S^1 \to S^1$ (both summands in orientation-preserving way) is evidently homotopic to a constant map. Looking on $H_1$ we have $\mathbb{Z} \xrightarrow{f_*} \mathbb{Z}\oplus \mathbb{Z} \xrightarrow{p_*} \mathbb{Z}$.
$p_*$ is addition: $S^1 \xrightarrow{i_1} S^1\vee S^1 \xrightarrow{p} S^1$ is identity (similary for $i_2$). $i_1$ induces $\mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z}$, to make sure it is an inclusion of the first summand we may use naturality of exact sequences for a map $(S^1, *) \xrightarrow{i_1} (S^1\vee S^1, \mathrm{im}\, i_2)$. Now we know $p_*(1, 0) = 1$ and $p_*(0, 1) = 1$, by linearity $p_*$ is addition.
Thus $f_*$ maps $1\in H_1(S^1)\cong \mathbb{Z}$ to some pair $(x, -x)\in H_1(S^1\vee S^1)$. To find out what $x$ is, consider the composition $S^1 \xrightarrow{f} S^1 \vee S^1 \to S^1$ where the second map collapses second summand to a point (it induces projection on a second coordinate in $\mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}$ too, e.g. by naturality of exact sequences of pairs). By construction this map is homotopic to identity. Similary for the second summand in $S^1\vee S^1$ we get that orientation-reversing map induces $-1$ on homology.
That's quite cumbersome to read, but actually very straightforward.
UPDATE
As mentioned in comments, here is proof reversing orientation induces identity on $H_0$. $* \to S^1$ and $*\to S^1 \to S^1$ (the second map is reversal of orientation) are equal as maps from a point, thus induce the same map on $H_0$. If we know that the inclusion of a base point in $S^1$ induces non-zero map on $H_0$, we're done, since automorphism of $H_0(S^1) \cong \mathbb{Z}$ fixing a non-zero element must be identity. To see this, use the naturality of long exact sequences of pairs for a map $(S^1, *) \to (*, *)$. The first commutative square (with non-relative $H_0$s) consists of three identity mapping of a point and a map induced from inclusion of a base point, and cannot commute unless $H_0(*) \to H_0(S^1)$ is an inclusion.
(Reply to a comment: nullhomotopic map induces zero on $H_1$ since it factors as map $X \to * \to Y$)
Also it seems to me that $X \hookrightarrow X \vee Y$ induces inclusion to the first summand in homology not due to functoriality properties of long exact sequences but this is just the way we identify $H_i(X\vee Y)$ with $H_i(X)\oplus H_i(Y)$. That is, we look at the pair $(X \vee Y, Y)$, write long exact sequence, then apply excision to get rid of relative homology and get $H_i(Y)$ instead. After that, we note that a lot of arrows in long exact sequence admit splitting and define the isomorphism $H_i(X \vee Y) \to H_i(X) \oplus H_i(Y)$ the way inclusion of $X$ in a wedge sum is inclusion on first summand. Thus some arguments by naturality above are not needed.
Best Answer
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.