First of all, let me point out that in the last section of the exact sequence, you're dealing with pointed sets and not groups. In particular, from $*\to \pi_1(\mathbb RP^2)\to \pi_0(\mathbb Z/2)\to *$ being an exact sequence of pointed sets, you can't conclude as easily that $\pi_1(\mathbb RP^2)\cong \mathbb Z/2$.
For instance, for any nontrivial group $G$, there is an exact sequence of pointed sets $*\to G\to \{0,1\}\to *$. So you have to put in a little more work if you only want to use that exact sequence (the usual road to computing this $\pi_1$ is through covering theory; but if you use additional data, I think you can use the exact sequence as well)
Your computations for $\times$ and $\vee$ from the computation of $\pi_1(\mathbb RP^2)$ are correct.
$\pi_i(\mathbb Z/2) = 0$ for $i>0$ because each point in $\mathbb Z/2$ is, well, a point, so it has trivial homotopy groups. You should try to prove that.
As pointed out by William in the comments, you could use covering theory to compute $\pi_1(\mathbb RP^2\vee \mathbb RP^2)$ although I wouldn't be surprised if that turned out to be a mess (just look at the result you're supposed to get). Van Kampen is the usual road for this computation (and seeing the result, it's reasonable to expect it to be the only sensible road)
How can $A$ and $B$ be $\mathbb{RP}^2$ if we only identify points on the equator of $S^2$ with their antipodal points?
Let $S^2$ be the sphere as usual. Now, cut $S^2$ along its equator gives us 2 $D^2$ (disks). Since $\sim$ is generated by $x\sim -x$ for all $x$ on the equator (i.e. $x$ is equivalent to its antipodal point iff its on the equator). We finally obtain two identical quotient space $A\approx D^2/{\sim}$ and $B\approx D^2/{\sim}$ where the same equivalence relation in the restricted situation identifies $x$ with $-x$ if $x\in\partial D^2\approx S^1$. Now, you can easily prove that $A\approx B\approx \Bbb{RP}^2$.
Remark: I think you confused the definition of $\Bbb{RP}^2$ with $A,B$. Indeed, $\Bbb{RP}^2=S^2/(x\sim-x)$. But, after identifying the interior of the upper hemisphere with the interior of the lower hemisphere, you'll get $D^2/(x\sim-x)$ where $x\in\partial D^2$ which is just what your instructor mentioned in the first paragraph.
How exactly do I find the normal subgroup $N$?
First, note that $X\not\approx\Bbb{RP}^2$ but $X$ is obtained by attaching $\partial(I^2/{\sim})$ where $(x,0)\sim (1-x,1)$ and $(0,y)\sim(1,1-y)$ to another identical copy of it (i.e. identifying the boundary of two $D^2/{\sim}$ together). Here is what it's like:
Let $U$ represent the blue half which is $\Bbb{RP}^2$ and $V$ be the other half. Clearly, $\pi_1(A\cap B)\cong \pi_1(S^1)=\langle a\rangle$. By Van-Kampen's Thm, $j_1:(U,x_0)\to(X,x_0)$ and $j_2:(V,x_0)\to(X,x_0)$ induce an epimorphism $j_*:\pi_1(U)*\pi_1(V)\to\pi_1(X)$. The amalgamated relation which is $N$ must be given by $i_{1*}:\pi_1(U\cap V)\to\pi_1(U)$ and $i_{2*}:\pi_1(U\cap V)\to\pi_1(V)$ so they have the form of $i_{1*}(a)^{-1}i_{2*}(a)=1$ where $a$ is a generator of $\pi_1(U\cap V)$. To see this, You can easily show that $N\subset \ker(j_*)$ using the commutativity of the diagram of this theorem, that is splitting $U\cap V\to X$ into two branches. Then, show the injectivity of $\pi_1(U)*\pi_1(V)/N\to\pi_1(X)$ which should be included in the proof of Seifert Van-Kampen's Thm.
We get $\pi_1(X)\cong(\pi_1(\Bbb{RP}^2)*\pi_1(\Bbb{RP}^2))/(aa=1,bb=1)$, but
since the generators of the two parts are the same (by the equivalence relation), we simplify the result to be $\pi_1(X)=\langle a,b\mid a=b,aa=1,bb=1\rangle=\langle a\mid aa=1\rangle\cong\Bbb{Z}/2$. Here $aa=a^2=1$ because in the picture $a^2\simeq$ (the outer boundary of the square and is a trivial loop).
Compute $H_*(X)$:
I prefer Cellular homology in this case because $X$ is just $\Bbb{RP}^1$ with 2 $e^2$ attached by gluing their boundaries with $\Bbb{RP}^1$.
- $H_0(X)\cong\Bbb{Z}$ because $X$ is connected. $H_p(X)=0$ if $p>2$.
For other cases, consider the cellular chain complex:
$$0\to\Bbb{Z}\oplus\Bbb{Z}\overset{\partial_2}{\to}\Bbb{Z}\overset{0}{\to}\Bbb{Z}\to0$$
For $H_2(X)$, Let's assign a counterclockwise orientation to 2-cells $\gamma_1,\gamma_2$, then we can see that $\partial_2(n\gamma_1+k\gamma_2)=0$ iff $n=k\implies Z_2(X)=\ker(\partial_2)=\Bbb{Z}$ because $\partial(\gamma_1)=2e^1$, $\partial(\gamma_2)=-2e^1$. So, $H_2(X)\cong\Bbb{Z}$
For $H_1(X)$, the image of the boundary map $D_1(X)\to D_0(X)$ is $0$ because the only 1-cell under the boundary map gives us $v_0-v_0=0$ which implies $Z_1(X)=\Bbb{Z}$ and from number 2, we know that $B_1(X)=im(\partial_2)=2\Bbb{Z}$ because $\partial_2(\gamma_1)$ wraps the equator two times and so does $\gamma_2$ (different directions). So, $H_1(X)=Z_1(X)/B_1(X)=\Bbb{Z}/2$.
To sum up,
$$
H_p(X;\Bbb{Z})=
\begin{cases}
\Bbb{Z} & p=0,2\\
\Bbb{Z}/2 & p=1\\
0 & \text{otherwise }
\end{cases}
$$
I think that Using MV sequence is a little bit complicated in this case, but you can check out this which is a similar situation and that person uses MV sequence which I think it's more difficult than Cellular homology.
If this helps you solve the problem, please consider to click the tick to accept it because it took me some time to write it in Latex format...
Best Answer
I think the question meant to ask about $\pi_2(S^2 \vee S^2)$. Since $\pi_0$ and $\pi_1$ vanish, the Hurewicz map gives an isomorphism $\pi_2(S^2 \vee S^2) \to H_2(S^2 \vee S^2) = \mathbb{Z}^2.$