Let's think about what the equivalence relation we put on points of $S^n$ is. We identify antipodal points, which means that the fiber of a point in $\mathbb{R}P^n$ under the quotient map is just two opposite points. Similarly, if you look at the lift of a neighborhood in $\mathbb{R}P^n$, as long as it is small enough, it's just going to be two copies of that neighborhood on opposite ends of the sphere.
In dimensions one and two, we can quickly come up with a model for $\mathbb{R}P^n$ that we can put down on paper and stare at. For $n=1$, we start with the circle $S^1$. Our equivalence relation is to identify opposite points, so let's find a set of representative points on $S^1$ that are in bijection with points in $\mathbb{R}P^1$. Any line through the center of the circle will intersect at least one point in the upper half of the circle. Furthermore, the only line that intersects it in two places is the horizontal line. So this means that a set of representatives can be taken to be those $(x,y)$ on the circle with $y > 0$, along with one of the points $(1,0)$ and $(-1,0)$; it doesn't matter which.
So now we have $\mathbb{R}P^1$ as a set, and we want to make sure it has the right topology. A neighborhood of our exceptional point $(1,0)$ is "missing" some neighbors: it is also close to points of the form $(x,y)$ with $x > 0$ and $y< 0$ small. But these points get identified with $(-x,-y)$, which are just points on the other end of the upper half-circle. So this tells us that we need to "glue" together $(1,0)$ and $(-1,0)$. In our construction of $\mathbb{R}P^1$, we started with a half-open line segment, and then glued the ends together, and this is nothing more than a circle.
As for your question about fundamental groups (incidentally, the standard notation is to use a lowercase $\pi$), we only need to look at what happens to a generator of $\pi_1(S^1)$, the loop that goes around once. Since we built $\mathbb{R}P^1$ from $S^1$ essentially by shrinking our circle by a factor of two, we see that the loop that goes around the big circle $S^1$ once will go around the smaller circle $\mathbb{R}P^1$ twice.
You should think about what $\mathbb{R}P^2$ "looks like", as you can do an analogous construction to get a model for $\mathbb{R}P^2$ from $S^2$ like I did for $S^1$. What happens at the boundary gets more interesting, but it's something to think about.
A covering transformation is in particular an element of ${\rm Homeo}(S^1\times\mathbb{RP}^2)$, so it induces an automorphism of homology $\varphi\in {\rm Aut}(\mathbb Z\times \mathbb Z_2)$. Now it is an easy exercise to show that $ {\rm Aut}(\mathbb Z\times \mathbb Z_2)$ consists of the identity map, $(a,b)\mapsto (-a,b)$, $(a,b)\mapsto (-a,b+a)$ and $(a,b)\mapsto (a,b+a)$. So we just have to rule these last three maps out. Suppose $f$ is a covering transformation inducing one of these maps. Now we use the fact that covering transformations cannot have fixed points and the Lefschetz fixed point theorem comes to the rescue. In the two maps where the coefficient of $a$ is negative, the alternating sum of traces of $f$ on the rational homologies $H_0\cong \mathbb Q$ and $H_1\cong\mathbb Q$ is $1-(-1)\neq 0$. So $f$ has a fixed point, which contradicts the possibility that it is a covering transformation. Hence we only need to rule out $(a,b)\mapsto (a,b+a)$. Unfortunately I don't see how to do that at the moment...
I'd like to add that this question depends on the particular space $S^1\times\mathbb{RP}^2$ and not just its fundamental group, as the following example shows. Let
$$G=\langle a,b,c|[a,b]=1, b^2=1, cac^{-1}=ab, [c,b]=1\rangle.$$
The subgroup generated by $a,b$ is a normal subgroup. It is clear that it is a quotient of $\mathbb Z\oplus\mathbb Z_2$, but in fact you can show that it is isomorphic to this group (see below). So let $X$ be a space with $\pi_1(X)=G$, and consider the normal cover $Y$ with fundamental group $\mathbb Z\oplus\mathbb Z_2$ corresponding to this subgroup. Then the deck transformation group is isomorphic to $\mathbb Z$ generated by $c$ and it acts on $H_1(Y)$ by sending $a$ to $a+b$ and $b\mapsto b$.
To see that the subgroup generated by $a$ and $b$ is isomorphic to $\mathbb Z\oplus\mathbb Z_2$, we give a different description of $G$. Notice that $b$ commutes with everything and $ca=acb$, so in a word representing an element in $G$, we can always rearrange it to be of the form $a^kc^\ell b^m$, with $a^kc^\ell b^m\cdot a^uc^v b^w=a^{k+u}c^{\ell+v}b^{m+w+\ell u}$. So $G$ is isomorphic to $\mathbb Z\times\mathbb Z\times \mathbb Z_2$ with a funny group multiplication: $(x,y,z)*(x',y',z')=(x+x',y+y',z+z'+y\cdot x').$ One can see that $\mathbb Z\times\{0\}\times \mathbb Z_2$ forms a subgroup isomorphic to $\mathbb Z\times\mathbb Z_2$.
Best Answer
As Jim pointed out, it's not enough to show that $f_*$ induces the trivial map on fundamental groups. To finish it, we can proceed as follows. Let $p\colon \mathbb R\to S^1$ be the covering projection. Since $f_*(\pi_1(P^2))=0$ (as Fredrik showed), it's clearly contained in $p_*(\pi_1(\mathbb R))=0$. So, this means there is a map $\tilde f\colon P^2\to\mathbb R$ such that $f=p\tilde f$ (Proposition 1.33 in Hatcher). Now, $\mathbb R$ is contractible so we have a homotopy $H\colon P^2\times I\to \mathbb R$ such that $H_0=\tilde f$ and $H_1$ is constant at some $x_0$. But then $pH$ is a homotopy from $p\tilde f=f$ to $p(x_0)$ meaning that $f$ is null-homotopic.