$H_0(X) = \mathbb{Z}$ since it's connected.
$H_2(X) = \mathbb{Z}^2$ since there are two non-contractable spheres
$H_1(X) = \mathbb{Z}$ since the torus has two non-contractible curves, but now one can be contracted along the sphere.
Let $a$ be the torus, and $b\cup b' = S^2$ be the upper and lower hemispheres.
$\partial a = a_1 + a_2 - a_1 - a_2 = 0$
Let $a_1 = \mathbb{T}^2 \cap S^2$ be the meridian of the sphere (as well as the torus)
$\partial b = a_1 =\partial b'$
Let $a_{12}$ be a point of $a_1 =\mathrm{torus}\cap \mathrm{sphere}$:
$\partial a_1 = \partial a_2 = \partial b_1 =a_{12} - a_{12}=0$
$H_0 = \ker\partial / \mathrm{im} \,\partial = \mathrm{span}\langle a_{12}\rangle = \mathbb{Z}$
$H_1 = \ker\partial / \mathrm{im} \,\partial = \mathrm{span}\langle a_1, a_2\rangle /
\mathrm{span}\langle a_1\rangle = \mathbb{Z}$
$H_2 = \ker\partial = \mathrm{span}\langle a,b-b'\rangle = \mathbb{Z}^2$
You may also observe the meridian circle $\mathrm{sphere} \cap \mathrm{torus}$ is contractible to a point.
In Hatcher Chapter 0, example 0.8 (page 12) shows us the sphere with two points identified is the homotopy equivalent to the wedge of a sphere and a circle.
$$ S^2 \simeq S^2 \vee S^1$$
For your example, $ \mathbb{T}^2 \cup_{S^1} S^2 \simeq S^2 \vee S^2 \vee S^1$.
Their homology groups will also be the same. $H_0 = \mathbb{Z}, H_1 = \mathbb{Z}, H_2 = \mathbb{Z}^2$.
Let $X$ be the first $(k - 1)$ factors in $A_k$ and $Y$ be the last factor. Let $S^1$ be the circle that connects $X$ with $Y$. Let $W$ be a neighborhood of $S^1$ that deformation retracts onto $S^1$. Define $U = X \cup W$, $V = Y \cup W$.
Using he Mayer-Vietoris sequence for reduced homology, we have the following exact sequence:
$$
\tilde H_1(U \cap V) \xrightarrow{\varphi} \tilde H_1(U) \oplus \tilde H_1(V) \xrightarrow{\psi} \tilde H_1(A_k) \rightarrow 0
$$
By considering the fundamental polygon for $\Bbb R \textrm P^2$, one can see that $U$ deformation retracts onto the wedge sum of $(k - 1)$ circles. Similarly, $V$ deformation retracts onto a circle. Hence $\tilde H_1(U) \cong \Bbb Z^{k-1}$, $\tilde H_1(V) \cong \Bbb Z$. Since $U \cap V = W$ deformation retracts onto $S^1$, we have $\tilde H_1(U \cap V) \cong \Bbb Z$.
What's left is to find $\ker \psi = \operatorname{im} \varphi$ and apply the first isomorphism theorem.
From the definition of the Mayer-Vietoris sequence, $\varphi = (i_*, j_*)$ where $i : U \cap V \hookrightarrow U$, $j : U \cap V \hookrightarrow V$ are the inclusion maps.
Since the generator of $\tilde H_1(U \cap V)$ goes twice around each generator of the circles of $U$, we have
$$
i_*(1) = \underbrace{(2, \ldots, 2)}_{(k - 1) \text{ times}}.
$$
Similarly, $j_*(1) = 2$.
Hence
$$
\ker \psi \cong \underbrace{(2, \ldots, 2)}_{k \text{ times}} \Bbb Z.$$
By an application of the first isomorphism theorem, we have
$$
\tilde H_1(A_k) \cong \left(\tilde H_1(U) \oplus \tilde H_1(V)\right) / \ker \psi = \Bbb Z^{k-1} \oplus \Bbb Z_2.
$$
Note that one can prove the following more general result:
If $M_1$ and $M_2$ are closed manifolds then there are isomorphisms $H_i(M_1 \# M_2) \cong H_i(M_1) \oplus H_i(M_2)$ for $0 < i < n$, with one exception: If both $M_1$ and $M_2$ are nonorientable, then $H_{n−1}(M_1 \# M_2)$ is obtained from $H_{n−1}(M_1) \oplus H_{n−1}(M_2)$ by replacing one of the two $\Bbb Z_2$ summands by a $\Bbb Z$ summand.
The proof is similar to what I have above, but requires some manifold theory.
Best Answer
You'll want to use the fact that $\mathbb{R}P^n$ can be written as $\mathbb{R}P^{n-1}\cup_f D^n$ where $D^n$ is the $n$-dimensional ball, and $f\colon S^{n-1}\to\mathbb{R}P^{n-1}$ is a 2-fold covering map, so we are gluing the $n$-ball along its boundary to $\mathbb{R}P^{n-1}$ via this map. You can then use Mayer-Vietoris and induced maps to explicitly work out the connecting map. In your case, you have $\mathbb{R}P^2=M\cup_f D^2$, where $M$ is the Mobius strip and $f\colon S^1\to M$ is the doubling map up to homotopy, or just the inclusion of the boundary into the Mobius strip.
To be more explicit, via Mayer-Vietoris, we get a long exact sequence $$\cdots\to H_2(M)\oplus H_2(D^2)\to H_2(\mathbb{R}P^2)\to H_1(S^1)\to H_1(M)\oplus H_1(D^2)\to\cdots$$ which, using the fact that $H_2(M)=H_2(D^2)=H_1(D^2)=0$ and $H_1(S^1) \cong H_1(M) \cong\mathbb{Z}$, reduces to the exact sequence. $$\cdots\to 0\to H_2(\mathbb{R}P^2) \stackrel{g}{\to} \mathbb{Z} \stackrel{\times 2}{\to} \mathbb{Z} \to\cdots$$
where we get that $\times 2$ map in the above sequence from the fact that the inclusion of the intersection of the two spaces (homotopy equivalent to a circle) into the Mobius strip is (up to homotopy) the degree-$2$ covering map, which induces multiplcation by $2$ in first homology.
By exactness, the image of $g$ must be $0\subset\mathbb{Z}$ as the doubling map in injective, but $g$ must itself be injective by exactness because the map $0\to H_2(\mathbb{R}P^2)$ has trivial image. The only way both of these conditions on $g$ can be satisfied is if $H_2(\mathbb{R}P^2)$ is trivial.