Does the tangent bundle on a 2-sphere span $\mathbb R^3$ and how are the operations defined? As a follow-up to this question I would like to clarify whether the tangent bundle on a sphere in $\mathbb R^3$ spans $\mathbb R^3$ to make sure I get the concept.
The tangent bundle is the set of the tangent planes at every single point on the surface of the 2-sphere $S$ and would be defined as
$$TS := \{(p, v):\ p\in S, v\in T_pS\}$$
If I get the idea correctly, there would be tangent planes through each point on the surface like the following ones in the drawing representing 3 single points ($P, S, Q)$:

Each plane would be translated to go through the origin to construct a vector space of tangent planes:

If the above is correct, the intuition is clear: the tangent bundle on the sphere would enable us to find a plane in any possible orientation, and hence, the disjoint union of these tangent planes would span $\mathbb R^3.$ Or is the disjoint piece a game changer?
QUESTIONS:

  
*
  
*Does this "fan" of translated tangent planes span $\mathbb R^3$?
  
*And how are the addition and scalar multiplication of a vector space defined on this tangent bundle?
  

 A: The tangent bundle $TS$ of the two dimensional sphere $S$ is the set you mentioned in the question.  I'm not sure in what sense you ask if it spans $\mathbb R^3$ though ($TS$ is not even a subset of $\mathbb R^3$). The tangent bundle does not, in general, have the structure of a vector space. Each tangent space $T_pS$ can be given one, but their disjoint union can't.
The fact that $TS$ is the disjoint union of the $T_pS$ and not simply their union is key. You want to keep track of the point the vectors in the tangent space are tangent to in order to keep vectors coming from different tangent spaces separate. So when you form the tangent bundle you prefix every tangent vector with the point it's tangent to, from which the definition $TS=\{(p, v):\ p\in S, v\in T_pS\}$. Since the manifold you are starting from, $S$, is a $2$-dimensional object, and you are attaching $2$-dimensional objects $T_pS$ at every point $p\in S$, it's reasonable to think that $TS$ should be a $4$-dimensional object. And in fact it is, it's a $4$ dimensional manifold, but not, in general, a $4$-dimensional vector space. 
That is generally speaking because, in some cases, like $T\mathbb R^n$, there is the structure of a $2n$ dimensional vector space since $T\mathbb R^n$ is essentially $\mathbb R^n\times \mathbb R^n$, but in general there is no meaningful vector space structure on a tangent bundle.
Note on the disjoint union: the disjoint union is a general construction. If for every element $a$ in a set $A$ you have another set $B_a$ (that may depend on $a$), the disjoint union is
$$
  \biguplus_{a\in A} B_a:=\bigcup_{a\in A} \{a\}\times B_a = \{(a, b):\ a\in A, b\in B_a\}
$$
