The following approach has the benefit that it is elementary in the sense that it only uses basic (homological and linear) algebra and the tools you allude to. More on this later*. I will just expand on the relevant parts of what is in this link.
You can use the antipode $A:S^n \to S^n$ to decompose $\Omega^k(S^n)$ in a direct sum of two subspaces $\Omega^k(S^n)_+ \oplus \Omega^k(S^n)_-$, since the induced map $A^*$ is such that $A^*{}^2=\mathrm{Id}$ (the subspaces are the eigenspaces of $1$ and $-1$, respectively). By naturality of $A^*$, $d$ respects that decomposition. It follows that $H^k(S^n) \simeq H^k(\Omega(S^n)_+)\oplus H^k(\Omega(S^n)_-)$ (the right side is "algebraic" cohomology originated from the complex $\Omega(S^n)_+$, resp. $\Omega(S^n)_-$), and it is clear that each one is the eigenspace of $A^*$ (on the cohomology level) of eigenvalue $+1$ and $-1$, respectively.
Since $A$ has degree $(-1)^{n+1}$ (it is a diffeomorphism that reverses/preserves orientation accordingly with the parity of $n$) and $H^n(S^n;\mathbb{R})=\mathbb{R}$, we have that $H^n(S^n)=H^n(\Omega(S^n))_+$ if $n$ is odd, and $H^n(S^n)=H^n(\Omega(S^n))_-$ if $n$ is even.
If we show that $\pi^*: \Omega^k(\mathbb{R}P^n) \to \Omega^k(S^n)_+$ is a isomorphism (the fact that the image indeed lies on $\Omega^k(S^n)_+$ follows from the fact that $\pi A=\pi$, because then $A^*\pi^*=\pi^*$), since it is natural it will follow that $H^k(\mathbb{R}P^n) \simeq H^k(\Omega(S^n))_+$, and hence we will have calculated the de Rham cohomology of $\mathbb{R}P^n$, since we know who are the $H^k(\Omega(S^n))_+$ (this is a good moment to see if you are following up: what is $H^k(\Omega(S^n))_+$, in terms of $n,k$? The only one to think about really is when $n=k$, the rest must be trivial - except $H^0$, of course).
To show that it is an isomorphism, you can check in the pdf I linked or use the answers by Mariano and user432847 here, which deal with a more general statement. Maybe seeing both would be best, since the general outlook that Mariano gives makes the fact that $\pi^*$ is an isomorphism with $\Omega^k(S^n)_+$ more intuitive.
*There are other methods to compute the cohomology of the projective space. Cellular homology + universal coefficients + deRham theorem is one example which you may want to check out.
Let me address items 1 and 3 in a cetain amount of detail, and item 2 very briefly at the end.
You say you are aware of homotopy equivalences and Mayer-Vietoris, but don't neglect homeomorphisms!
First, from what you are aware regarding $\mathbb R^3 \setminus \{\text{a single line}\}$ I'm sure you could figure out that $\mathbb R^3 \setminus \{\text{$n$ parallel lines all in the same plane}\}$ is homotopy equivalent to $\mathbb R^2 \setminus \{\text{$n$ points all in the same line}\}$.
The key idea is to prove:
The homeomorphism type of $\mathbb R^3 \setminus \{\text{$n$ pairwise disjoint lines}\}$ is well-defined, depending only on $n$.
So in your case 1, the space is homeomorphic to $\mathbb R^3 \setminus \{\text{two disjoint, parallel lines}\}$ which is homotopy equivalent to $\mathbb R^2 \setminus \{\text{two points}\}$.
Let me do the case $n=2$ in full detail, after which I'll suggest some ideas for going beyond that case.
Lemma: $R^3 \setminus \{\text{two disjoint non-parallel lines}\}$ is homeomorphic to $\mathbb R^3 \setminus \{\text{two disjoint parallel lines}\}$.
Consider two disjoint non-parallel lines $L,M$. Connect them by the unique line segment $\overline{A B}$, $A \in L$, $B \in M$ that means $L$ and $M$ at right angles. Let $P$ be the plane that bisects $A,B$ perpendicularly. By applying a rigid motion of Euclidean space, we may assume that $P$ is the $xy$-plane, $A = (0,0,c)$, and $B = (0,0,-c)$. The line $L$ lies in the plane $z=c$, and by applying a rigid rotation of Euclidean space around the $z$-axis we may assume that $L$ is parallel to the $x$-axis. Looking down the $z$-axis and projection to the $x,y$ plane, the projection of $L$ is the $x$-axis, and the projection of $M$ makes some positive angle $\theta$ with respect to the $x$-axis.
Now we are going to apply a non rigid homeomorphism of Euclidean 3-space, in three parts.
First, all points with $z \ge 0$ are fixed.
Second, the entire portion of the plane with $z \le -c$, is rigidly rotated around the $z$ axis by an angle of $-\theta$; after this rotation, the image $M'$ of the line $M$ is parallel to the $x$-axis hence to $L$.
Third, for $-c \le z_0 \le 0$, the plane $z=z_0$ is rotated rigidly around the $z$ axis by an angle $-\theta \cdot \frac{z_0}{c}$.
That's it for $n=2$!
I think the best thing to do in general is to use the ideas of this proof to construct a proof by induction for larger values of $n$.
For some final comments regarding item 2, instead of planes, think of spheres centered at the intersection point $P$: the space $\mathbb R^3 - \{\text{two lines intersecting at $P$}\}$ is homotopy equivalent to a sphere centered on $P$ minus the four points where those lines puncture that sphere.
Best Answer
If you remove a single point from $S^2$, the remaining space is homeomorphic to $\mathbb{R}^2$. If you remove more than one point, the homeomorphism will map the additional points to points in $\mathbb{R}^2$. So yes, a sphere with $k$ points removed is homeomorphic to $\mathbb{R}^2$ with $k-1$ points removed.