Computation of Pushforward and Pullback After reading about the pushforward and pullback, I don't really have a concrete grasp of them, so I think these simple questions might clear things up for me; I appreciate any hints or solutions.  
Let $(s,t)$ be coordinates on $\mathbb{R}^2$ and $(x,y,z)$ be coordinates on $\mathbb{R}^3$.  Let $g:\mathbb{R}^2\to\mathbb{R}^3$ be defined by $g(s,t)=(\sin(t),st^2,s^3-1)$.
Let $X_p\in T_p\mathbb{R}^2$ be given by $X_p=\frac{\partial}{\partial s}|_p - \frac{\partial}{\partial t}|_p$, compute the push-forward  $g_*X_p$.
Let $\omega$ be the smooth $1$-form $\omega=dx+xdy+y^2dz$; find the pullback $g^*\omega$.
From what I understand, if $\varphi:M\to N$ then $d\varphi_p:T_p M\to T_{\varphi(p)}N$ and is given by $d\varphi_p(X)(f)=X(f\circ \varphi)$ where $X\in TM (T_pM?)$ and $f\in C^\infty (N)$.  I think that $d\varphi=\varphi_*$?  And I reckon that in this case $M=\mathbb{R}^2$,  $N=\mathbb{R}^3$, and $\varphi=g$ but I'm not sure what $f$ equals (what does $g_*X_p$ act on?).... do we even need to act on an $f$ to find $g_*X$?  Sorry I'm a bit lost.
 A: Let's work out some computations: first, the push-forward of the vector field $X = \frac{\partial }{\partial s} - \frac{\partial }{\partial t}$, denoted $g_* X$. To do so, we need to find the push forward of the two vector fields $\frac{\partial}{\partial s}$ and $\frac{\partial}{\partial t}$. This is given by
$$
g_* \left( \frac{\partial}{\partial s} \right)
= \frac{\partial x}{\partial s} \frac{\partial}{\partial x} + \frac{\partial y}{\partial s} \frac{\partial}{\partial y} + \frac{\partial z}{\partial s}\frac{\partial }{\partial z} =  t^2 \frac{\partial}{\partial y} + 3s^2 \frac{\partial}{\partial z},
$$
and
$$
g_* \left( \frac{\partial}{\partial t} \right)
= \frac{\partial x}{\partial t} \frac{\partial}{\partial x} + \frac{\partial y}{\partial t}\frac{\partial}{\partial y} + \frac{\partial z}{\partial t} \frac{\partial}{\partial z} = \cos(t) \frac{\partial}{\partial x} + 2st \frac{\partial}{\partial y}.
$$
Then,
$$
g_* X = g_* \left( \frac{\partial}{\partial s} \right) - g_* \left( \frac{\partial}{\partial t} \right) = -\cos(t) \frac{\partial}{\partial x} + (t^2 -2st) \frac{\partial}{\partial y} + 3s^2 \frac{\partial}{\partial z}.
$$
A remark: in $g_* X$, I should have written this without reference to $s$ and $t$ (that is, solve the equations $x=\sin(t)$ and $y=st^2$ and $z=s^3-1$ for $s$ and $t$ in terms of $x,y,z$), however it was not clear to me how to do that here.
Next, let's find the pullback $g^{*}\omega$: this will be a 1-form on $\mathbb{R}^2$ and hence should be of the form $a(s,t) ds + b(s,t) dt$. To figure out what these functions $a(s,t)$ and $b(s,t)$ are, let us rewrite each `component' of the form $\omega$ in terms of $s$ and $t$:
$$
dx = \frac{\partial x}{\partial s} ds + \frac{\partial x}{\partial t} dt
= \cos(t) dt
$$
$$
dy = \frac{\partial y}{\partial s} ds + \frac{\partial y}{\partial t} dt
= t^2 ds + 2st dt
$$
$$
dz = \frac{\partial z}{\partial s} ds + \frac{\partial z}{\partial t} dt
= 3s^2 ds
$$
Putting all of this together, we find that
\begin{align*}
g^* \omega &= \cos(t) dt + \sin(t) (t^2 ds + 2st dt) + s^2 t^4 (3s^2 ds) \\
&= (t^2 \sin(t) + s^4 t^4 ) ds + (\cos(t) + 2st \sin(t) ) dt.
\end{align*}
If you find that you need practice with computations when dealing with manifolds and differential geometry (e.g. push forwards, pullbacks, Lie derivatives, etc.), I recommend looking at an exercise-heavy book such as Gadea & Masque's "Analysis and Algebra on Differentiable Manifolds, A Workbook for Students and Teachers". I found this one to be very helpful when learning this material for the first time.
