The lecturer answers exactly that question of a student five minutes later
"Because if you want to write just any vector field as a linear combination of this, then at the point where one of these vanishes, then at that point you will not be able to reproduce a component in that direction."
This argument proves that if any vector field necessarily vanishes at some point, then there is no basis for the module
of smooth vector fields with only $d$ elements, where $d$ is the dimension of $M$. In fact, if one of the vector fields vanishes at $P$, then the other $d-1$ supposed members of the
basis would span only a $d-1$ dimensional vector space inside $T_p M$. So any vector field $X$ with $X(P)$ outside this $d-1$-dimensional space could not be written as a linear combination of the $d$ vector fields.
However, this argument does not prove that there is no basis, there could be a basis with more than $d$ elements.
In order to have a complete proof, you must prove that the dimension cannot be greater than $d$. This is not so difficult either:
Assume that there are more than $d$ smooth vector fields $X_1,\dots,X_d,X_{d+1},\dots$ in the basis. Then at a given point $P$, they span all of $T_p M$ (otherwise the same argument as above gives a contradiction).
Then there is a subset of $d$ of the given vector fields, which we can assume (by renumbering) to be $X_1,\dots,X_d$, for which
$\{X_1(P),\dots,X_d(P)\}$ is a base (because $T_p M$ is a vector space of dimension $d$). So $X_1(P),\dots,X_d(P)$ are linearly independent, and
it is also true that there is a neighborhood $U$ of $P$ such that $X_1(Q),\dots,X_d(Q)$ are linearly independent for each $Q\in U$.
Take a bump function $g:M\to \Bbb{R}$, i.e. such that $g(m)=0$ for $m\notin U$ but $g(P)\ne 0$. Then
$g X_{d+1}$ can be written as a $C^{\infty}(M)$-linear combination of $X_1,\dots,X_d$, so
$X_1,\dots,X_d,X_{d+1}$ are not $C^{\infty}(M)$-linearly independent.
No, $N$ is never semisimple if $M$ is positive-dimensional. There are many ways to see this. For instance, if $f\in C^\infty(M)$ is any function supported on a single coordinate chart of $M$, then the localization $N_f$ is free over $C^\infty(M)_f$ of rank $\dim M$. If $N$ were semisimple, then $N_f$ would also be semisimple, and thus $C^\infty(M)_f$ would be semisimple if $\dim M>0$. But a semisimple ring is in particular Noetherian, and $C^\infty(M)_f$ is not Noetherian if $\dim M>0$ and $f$ is nonzero (for instance, any decreasing sequence of closed subsets on which $f$ is nonzero gives an increasing sequence of ideals in $C^\infty(M)_f$).
Best Answer
The normal bundle of $S^2$ embedded in $\mathbb{R}^3$ is trivial (see this question), so the sections of this bundle give a rank $1$ free $C^{\infty}S^2$ module $F$ so that $\mathfrak X(S^2) \oplus F= \oplus_{i=1}^3 C^{\infty}S^2$.