It turns out that the homology groups $H_{\bullet}(X,\mathbb{Z})$ of a space $X$ with $\mathbb{Z}$-coefficients determine the homology groups $H_{\bullet}(X,A)$ with coefficients in any abelian group $A$. The key result here is the (very well named!) Universal Coefficient Theorem.
The basic idea is that our first guess at $H_i(X,A)$ is simply $H_i(X,\mathbb{Z}) \otimes A$. This is a good first guess in that in all cases there is a natural injective map $H_i(X,\mathbb{Z}) \otimes A \rightarrow H_i(X,A)$. As you have seen, this map need not be an isomorphism. The universal coefficient theorem tells you that its cokernel is $\operatorname{Tor}(H_{i-1}(X,\mathbb{Z}),A)$ and also that the sequence is (non-canonically) split, i.e.,
$$H_i(X,A) \cong (H_i(X,\mathbb{Z}) \otimes A) \oplus \operatorname{Tor}(H_{i-1}(X,\mathbb{Z}),A).$$
Here $\operatorname{Tor}( \ , \ )$ is the first "Tor group" of homological algebra. It may well be that you don't know what this gadget is. (I didn't when I first learned algebraic topology.) So I found it helpful to write down a "cheatsheet" for $\operatorname{Tor}(X,Y)$ when $X$ and $Y$ are both finitely generated abelian groups. Indeed, since $\operatorname{Tor}$ is bi-additive and symmetric, it is enough to know that for all $m,n \in \mathbb{Z}^+$,
$\operatorname{Tor}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) = 0$ and
$\operatorname{Tor}(\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) \cong \mathbb{Z}/\operatorname{gcd}(m,n) \mathbb{Z}$.
As a first exercise, try to use all this information to confirm that the homology groups of $\mathbb{R} \mathbb{P}^2$ with $\mathbb{Z}/2\mathbb{Z}$-coefficients are as you said.
(There is also a Universal Coefficient Theorem for cohomology in which the correction term involves $\operatorname{Ext} = \operatorname{Ext}^1$ instead of $\operatorname{Tor}$...)
$\mathbb{R}^n$ is contractible therefore homotopy equivalent to a point and so $H_n(\mathbb{R}^m) = H_n(\{ \ast \})$.
$$ H_n(\{ \ast \}) = 0 , n > 0$$
$$ H_n(\{ \ast \}) = \mathbb{Z} , n = 0$$
Best Answer
View the real projective plane as the union of a disk $D^{2}$ and a Möbius strip $M$intersecting in an annulus, so that the edge of the disk goes all the way around the boundary of the Möbius strip. So when we consider $D^{2} \cap M \simeq S^{1}$, the circle "wraps around" the disk twice. Note that $\mathbb{R}P^{2}$ is a connected 2-dimensional manifold so $H_{0}(\mathbb{R}P^{2}) = \mathbb{Z}$ and $H_{n}((\mathbb{R}P^{2}) = 0$ for $n \geq 3$.
Now $M \simeq S^{1}$ and $M \cap D^{2} \simeq S^{1}$, the Mayer Vietrois sequence is
$$H_{2}(D) \oplus H_{2}(S^{1}) \longrightarrow H_{2}(\mathbb{R}P^{2})\longrightarrow H_{1}(S^{1}) \longrightarrow H_{1}(D) \oplus H_{1}(S^{1}) \longrightarrow H_{1}(\mathbb{R}P^{2}) \longrightarrow H_{0}(S^{1}) \longrightarrow H_{0}(D) \oplus H_{0}(S^{1}) \longrightarrow H_{0}(\mathbb{R}P^{2})\longrightarrow 0,$$ giving
$$0 \longrightarrow H_{2}(\mathbb{R}P^{2})\longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow H_{1}(\mathbb{R}P^{2}) \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow 0.$$
The map including the annulus into the Möbius strip wraps twice around the circle that is a retract of the Möbius strip, so the first map $\mathbb{Z} \rightarrow \mathbb{Z}$ is given by multiplication by $2$. Hence the map is injective and $H_{2}(\mathbb{R}P^{2}) = 0$. Now the last 3 groups in the sequence form a short exact sequence, so the map $H_{1}(\mathbb{R}P^{2}) \rightarrow \mathbb{Z}$ must be the zero map as $\mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ is injective. Hence we have a short exact sequence
$$ 0 \longrightarrow \mathbb{Z} \stackrel{2}{\longrightarrow} \mathbb{Z} \longrightarrow H_{1}(\mathbb{R}P^{2}) \longrightarrow 0,$$
so $H_{1}(\mathbb{R}P^{2}) = \mathbb{Z} / 2 \mathbb{Z}$.