Concrete example of vector field along a map In this book the definition of a vector field along a map $f: M \to N$ is given as follows:

I am currently trying to understand this definition. For this purpose I wanted to work out a concrete example. But I need some help. Here is the example:
Let $f: \mathbb R^2 \to \mathbb R^3$ be define by $f(t,s) = (t^2 + 2s, t^3 + 3ts, t^4 + 4t^2 s)$. 
To find a vector field along it I first calculated the derivatives $f_t$ and $f_s$ and then noticed that 
$$ f_t = t f_s + s (0,3,8t)$$
I then let $v_1 (t,s) := f_s$ and $v_2(t,s) := (0,3,8t)$.
And then I wanted to check if $\pi_{T\mathbb R^2} \circ v_i = f$ but this is where I am stuck at the moment. 
I believe that $\pi_{T\mathbb R^2} \circ v_1 = v_1 (t,s)$ and $\pi_{T\mathbb R^2} \circ
 v_2 = v_2( t,s)$. But by how we defined these vectors we would then have
$$ \pi_{T\mathbb R^2} \circ v_1 = v_1 (t,s) = f_s$$
when we want to have
$$ \pi_{T\mathbb R^2} \circ v_1  = f$$

Is $v_1$ really not a vector field along $f$ or am I missing
  something?

 A: It may be hard to see a lot from an example of the type you are considering. Keep in mind that a vector field simply associates to each point $x\in M$ a tangent vector $\nu(x)\in T_xM$ in a way that depends smoothly on $x$. Likewise, given $f:N\to M$, a vector field along $f$ just assigns to each $x\in N$ a tangent vector at the point $f(x)\in M$, i.e. $\nu(x)\in T_{f(x)}M$. Now if you choose $M=\mathbb R^3$, then $TM$ is naturally isomorphic to $M\times\mathbb R^3$. Hence given $f:N\to\mathbb R^3$ (for any $N$), a vector field along $f$ can be equivalently described by a smooth function $g:N\to\mathbb R^3$, by simply viewing $g(x)$ as a tangent vector at the point $f(x)$. This corresponds to the comment above, which says that one should actually write $(f,g):N\to\mathbb R^3\times\mathbb R^3\cong T\mathbb R^3$, to clarify the situation. Still, I don't think much interesting can be got out of such an example. 
There are several conceptual examples of vector fields along smooth functions. The simplest is the derivative $c'$ of a smooth curve $c:I\to M$ in a manifold, viewed as a smooth function $c':I\to TM$. A related example is to take the embedding $i:S^2\hookrightarrow\mathbb R^3$ of the unit sphere (and similarly of any other surface). Then associating to $x\in S^2$ the outward pointing unit normal at $i(x)$ defines a vector field along $i$. (For $S^2$, this is just $(i,i)$ as a map $S^2\times S^2\to\mathbb R^3\times\mathbb R^3\cong T\mathbb R^3$, but here it gets geometric content. In general, it recovers the Gauss map as the second component.) Finally, the "generic" example of a vector field along an arbitrary function $f:N\to M$ is taking a vector field $\nu\in\mathfrak X(N)$ and looking at $f_*\nu$ defined by $f_*\nu(x):=T_xf(\nu(x))$.     
A: I might as well make my comment into an answer. You make some slight mistakes, and the example you chose is a strange one: there are simpler examples.
(1) You should really write $v_1(s,t):=(f,f_s)$ to emphasize that $v_1$ takes values in $TM$, which has twice the dimension of $M$. (In other words, the points of $TM$ have the form $(p,v)$ for $p\in M$ and $v\in T_pM$.) You'll then see the projection condition works out. A vector field along a function attaches tangent vectors along base points in $M$ parametrized by the image of $f$.
You can look at the section starting at the bottom of page 55 in Lee's book to remind yourself of the definition of the tangent bundle and why we coordinatize it as $(p,v)$.
(2) $M$ is $\mathbb{R}^3$ in your example, so you care about the projection map $\pi$ on $T\mathbb{R}^3$, not $T\mathbb{R}^2$ as you wrote.
(3) You do not need to differentiate $f$ with respect to $s$ or to $t$ in order to construct a vector field along the map $f$. For other trivial examples, consider
$$v(s,t):=(f,(t,0,0))$$
or, still more trivial,
$$w(s,t):=(f,(0,0,0))$$
Perhaps the reason you took the partial derivatives of $f$ is that you meant to think of $f$ as a function not into $\mathbb{R}^3$ but into a surface $S$, namely a patch of the image of your function $f$. 
In that case you do need to be careful that the vector you attach to $f(p)$ lies in $T_{f(p)}S$; you can attach anything in the span of $f_t$ and $f_s$. But if that's what you wanted to do, you must first restrict the domain of $f$ so that your function is injective and so that the tangent vectors $f_s$ and $f_t$ are linearly independent.
