- On the vector bundle side, after discussing the cylinder and Möbius bundle and then moving to tangent bundles, you can explore the tangent bundles on $S^2$ and $T^2$ by looking at vector fields on them.
The tangent bundle of $T^2$ is trivial because there are two, linearly independent non-vanishing sections (forming a basis at every point) given by a vector field pointing in one direction around the torus and a vector field pointing in the other.
The tangent bundle of $S^2$, by contrast, has no non-vanishing sections, let alone two linearly independent ones (this is the hairy ball theorem), so the tangent bundle of $S^2$ is non-trivial.
- On the fiber bundle side, bundles over the circle are nice to visualize. You can see a lot of the bundle part happening without the picture becoming hard to visualize.
A circle bundle can be trivialized over all but one point, so all the action can be viewed in how you "glue" the bundle at that one point.
The cylinder and Möbius bundle are $\mathbb{R}$ bundles over the circle, where the the Möbius bundle is glued by the map $x \mapsto -x$ (and the cylinder is glued by the identity).
The torus and Klein bottle are circle bundles over the circle, where the torus is glued by the identity and the Klein bottle is glued by a map $S^1 \rightarrow S^1$ that "goes backwards", e.g. $\theta \mapsto -\theta$ (or alternatively, given by a reflection in $\mathbb{R}^2$ restricted to $S^1$).
- If you're extremely ambitious, you could draw 2-dimensional vector bundles over $S^2$ and discuss how they can be trivialized over the upper hemisphere and over the lower hemisphere. Then the bundle is determined by how you glue these together over the equator. If your class has done examples with atlases for $S^2$, you could then figure out what this gluing is in the case of $TS^2$.
A manifold is called parallelisable if it has trivial tangent bundle. So your question is for which positive integers $m$ is $\mathbb{RP}^m$ parallelisable? What about $\mathbb{CP}^m$?
If $T\mathbb{RP}^m$ is trivial, then the total Stiefel-Whitney class $w(T\mathbb{RP}^m)$ is $1$. By Corollary $4.6$ of Milnor & Stasheff's Characteristic Classes, $w(T\mathbb{RP}^m) = 1$ if and only if $m + 1$ is a power of $2$. So the only real projective spaces which can possibly be parallelisable are $\mathbb{RP}^1$, $\mathbb{RP}^3, \mathbb{RP}^7, \mathbb{RP}^{15}, \mathbb{RP}^{31}, \dots$ We still need to determine which of these are actually parallelisable (the condition on the total Stiefel-Whitney class is a necessary condition, but not sufficient as the case of spheres demonstrates).
Theorem $4.7$ of the same book states that if $\mathbb{R}^n$ admits a bilinear map $p : \mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}^n$ without zero divisors, then $\mathbb{RP}^{n-1}$ is parallelisable. By identifying $\mathbb{R}^2$ with $\mathbb{C}$, we obtain such a bilinear map for $n = 2$ given by the usual multiplication of complex numbers, i.e. $p((a, b),(c, d)) = (ac - bd, ac + bd)$. Similarly, by identifying $\mathbb{R}^4$ with the quaternions $\mathbb{H}$ and $\mathbb{R}^8$ with the octionions $\mathbb{O}$, we obtain such a map for $n = 4$ and $n = 8$. Therefore, $\mathbb{RP}^1$, $\mathbb{RP}^3$ and $\mathbb{RP}^7$ are parallelisable.
What about $\mathbb{RP}^{15}$, $\mathbb{RP}^{31}, \dots$? None of the remaining real projective spaces are parallelisable - this is harder to show. It follows from Kervaire's $1958$ paper Non-parallelizability of the $n$-sphere for $n > 7$.
So, the positive integers $m$ for which $\mathbb{RP}^m$ is parallelisable are $m = 1, 3,$ and $7$.
The parallelisability of $\mathbb{CP}^m$ is much easier to determine: $\mathbb{CP}^m$ is not parallelisable for any positive integer $m$.
As $\mathbb{CP}^m$ is a complex manifold, its tangent bundle is a complex vector bundle and hence has a total Chern class. The total Chern class of $\mathbb{CP}^m$ is $c(T\mathbb{CP}^m) = (1 + a)^{m+1} \in H^*(\mathbb{CP}^m, \mathbb{Z})$. If $T\mathbb{CP}^m$ were trivial, then it would have total Chern class $1$ but this is never the case. More explicitly,
$$c_1(T\mathbb{CP}^m) = (m+1)a \in H^2(\mathbb{CP}^m, \mathbb{Z}) \cong \mathbb{Z}.$$
As $a \neq 0$ and $m \neq -1$, $c_1(T\mathbb{CP}^m) = (m + 1)a \neq 0$.
More generally, any oriented closed manifold which is parallelisable must have Euler characteristic zero. In fact, such an oriented closed manifold which admits a nowhere zero vector field has Euler characteristic zero, see Property $9.7$ of the aforementioned book. The claim then follows from the fact that $\chi(\mathbb{CP}^m) = m + 1 \neq 0$.
Best Answer
If you want to describe tangent bundles, the appropriate language is classifying spaces.
A vector bundle $\mathbb R^k \to E \to B$ over a space $B$ is described by a homotopy-class of map
$$B \to Gr_{\infty,k}$$
where $Gr_{\infty,k}$ is the space of all $k$-dimensional vector subspaces of $\oplus_\infty \mathbb R$.
So for example, the tangent bundle of $S^2$ is a $2$-dimensional vector bundle over $S^2$, so described by a map
$$S^2 \to Gr_{\infty,2}$$
$Gr_{\infty,2}$ as a space would be called $B(O_2)$, the classifying space of the Lie group $O_2$, meaning that it is the quotient of a contractible space by a free action of $O_2$ (think of the associated Stiefel space). So an element of $\pi_2 Gr_{\infty,2}$ is equivalent (via the homotopy long exact sequence) to an element of $\pi_1 O_2$, which is isomorphic to $\mathbb Z$.
i.e. 2-dimensional vector bundles over $S^2$ are described by an integer.
There's another way to see the above construction. Decompose $S^2$ into the union of two discs, the upper and lower hemisphere. Via pull-backs this decomposes $TS^2$ into (up to an isomorphism) $D_u \times \mathbb R^2$ and $D_l \times \mathbb R^2$ where $D_u$ and $D_l$ are the upper and lower hemi-spheres respectively. $\partial D_u = \partial D_l = S^1$. So there's a gluing map construction
$$ TS^2 = (D_u \times \mathbb R^2) \cup (D_l \times \mathbb R^2) $$
There's is a map describing how point on $\partial D_l \times \mathbb R^2$ have to be glued to points on $\partial D_u \times \mathbb R^2$ and it has the form
$$(z,v) \longmapsto (z,f_z(v))$$
where
$$f : S^1 \to O_2$$
The homotopy-class of this map is again described by an integer. These are the same two integers. A fun calculation shows you it's two, the Euler characteristic.
The above story is worked-out in more detail in Steenrod's book on fibre bundles. Also Milnor and Stasheff.
By-the-way, many people have trouble initially thinking about tangent bundles. They're fairly delicate objects.