Question on vector fields along maps (need a quick sanity check) Let $f: M \to N$ be some smooth map between smooth manifolds. 
If $V$ is a vector field, that is, a smooth map $V: N \to TN$ then $V$ is a vector field along $f$ if the projection $\pi: TN \to N$ is such that $\pi (V(m)) = f(m)$ for all $m \in M$.
If ${\partial \over \partial x_i} f(m)$ is non zero then $V (m) := (f(m), {\partial \over \partial x_i} f(m))$ is a vector field along $f$ as $\pi (V(m)) = f(m)$.

Now my question is: what happens if ${\partial \over \partial x_i}
 f(m) = 0$? Can I choose $V$ to be anything other than $0$? It seems to
   me that the answer might be yes since the zero vector is part of any
  vector space.

I started to wonder about this as I was trying to come up with a vector field along a map $(f_1,f_2): \mathbb R \to \mathbb R^2$ mapping $0$ to $0$ and such that ${\partial^n \over \partial x^n}f_i(0) \neq 0$ for some $n \in \mathbb N_{>0}$. Furthermore, I want the vector field to be non zero and such that the linear span of the vector field contain the linear span of ${\partial \over \partial x}f(x)$ on some neighbourhood of $x_0 \in \mathbb R$.
 A: First, let's make sure we're clear about the definitions. A vector field on $N$ is a continuous map $V\colon N\to TN$ such that $V(n)\in T_nN$ for each $n\in N$. A vector field along $f$ is a continuous map $W \colon M \to TN$ such that $W(m) \in T_{f(m)}N$ for each $m\in M$. 
If $V$ is a vector field on $N$, then $V$ is not, as you claimed, a vector field along $f$; but it can be used to define a vector field along $f$ by setting $W=V\circ f$: In other words, $W(m) = V_{f(m)}\in T_{f(m)}N$ for each $m\in M$. But there are also plenty of vector fields along $f$ that are not of this type.
I'm not sure exactly what you mean by $\frac{\partial}{\partial x^i}f(m)$. Perhaps you mean to choose coordinates $(x^1,\dots,x^n)$ on some open subset $U\subseteq M$; if so, then $V(m) = \frac{\partial}{\partial x^i}f(m)$ does define a vector field along the restricted map $f|_U$. One way to make this precise is by letting $V(x^1,\dots,x^m)$ be the velocity vector at $t=0$ of the curve
$$
t\mapsto f(x^1,\dots, x^i+t,\dots, x^n).
$$
This is a well-defined vector field along $f|_U$, regardless of whether $\frac{\partial}{\partial x^i}f(m)=0$ for some $m$ or not.
To answer your last question, if $f = (f_1,f_2)\colon \mathbb R\to \mathbb R^2$ is a smooth map such that $f(0)=0$, and $n$ is the smallest positive integer such that $\frac{\partial^n}{\partial x^n}f(0)\ne 0$, then Taylor's theorem implies that $(f_1(x),f_2(x)) = t^n (W_1(x),W_2(x))$ for some smooth functions $W_1,W_2$ such that $W_1(0)$ and $W_2(0)$ are not both zero. You can think of $(W_1,W_2)$ as a vector field along $f$, and as such, it satisfies the condition you're interested in.
