Question about creating a volume form for $SL(2,\mathbb{R})$ This problem comes out of R.W.R. Darling (Differential Forms and Connections) ch.8. In the chapter he shows that if $M$ is an $n$-dimensional differential manifold immersed in $\mathbb{R}^{n+k}$, and $\Psi$ is an immersion from $\mathbb{R}^n \rightarrow \mathbb{R}^{n+k}$ that parametrizes the manifold, and $f$ is a submersion from $\mathbb{R}^{n+k} \rightarrow \mathbb{R}^k$ such that $f^{-1}(0) = M$, then we can construct a volume form $(\star df)$ on $M$ using the Hodge star, and that $(\star df)\Lambda^n \Psi_{*}$, which parametrizes the volume form, is given for $k=1$, and $\Psi(0) = r$ by 
$$ \begin{vmatrix} D_1 f(r) & D_1 \Psi_1(0) & \cdots & D_n \Psi_1(0) \\
                         \vdots & \vdots      & \ddots & \vdots      \\
                         D_n f(r) & D_1 \Psi_n(0) & \cdots & D_n \Psi_n(0) \\
\end{vmatrix}. $$
The exercise is to do this with the submanifold $SL(2,\mathbb{R}) \subset GL(2,\mathbb{R})$ regarded as equivalent to $\mathbb{R}^{nxn}$, with $\Psi$ parametrizing $\begin{pmatrix} x & y \\ z & w \\ \end{pmatrix}$ as the image of $(x,y,z)$ and $f(x,y,z,w) = xw - yz - 1.$ I calculated this, and got: 
$$ \begin{vmatrix} w & 1 & 0 & 0 \\
                  -z & 0 & 1 & 0 \\
                  -y & 0 & 0 & 1 \\
                   x & \frac{-w}{x} & \frac{z}{x} & \frac{y}{x} \\
\end{vmatrix}, $$
where $w = \frac{1+yz}{x}$, which correctly evaluates at $I$ to give $-2dx\land dy\land dz$ as the parametrized volume operator. 
The second part is where I have a problem, it says to extend this volume form in a left-invariant manner to $SL(2,\mathbb{R})$ by calculating $(L_A^{*}(\star df))(A^{-1})$, where $L_A$ is the left shift operator on $GL(2,\mathbb{R})$, with $L_A \begin{pmatrix} s & t \\ u & v \\ \end{pmatrix} = \begin{pmatrix} x & y\\ z & w\\ \end{pmatrix} \begin{pmatrix} s & t \\ u & v \\ \end{pmatrix}$ when $A = \begin{pmatrix} x & y\\ z & w\\ \end{pmatrix}.$
I sense that I should take $(L_A^{*}(\star df)) = ((\star df)L_{A{*}})$ to start the process, but am confused as to how the push-forward fits into the calculation with respect to the parametrization $\Psi$.  Could someone help me with how that works? And do I calculate $L_{A{*}}$ as an element of $\mathbb{R}^{nxn}$ which gives a differential as a 4x4 matrix? And if so, does it pre-multiply or post multiply which of the various pieces of the matrix determinant needed to form the volume form? (The problem is 8.4.5 in Darling, p.173, and this is a self-study question.)

First, I would suggest that you practice computing some pullbacks in a more basic setting. For example, if $f(u,v)=(u+v^2,uv,u^3+v^3)=(x,y,z)$ what is $f^*(x\,dy\wedge dz + z\,dx\wedge dy)$? You should learn how to do this without ever writing down a push-forward.
Second, you want to compute $L_{A^{-1}}^*(dx\wedge dy\wedge dz)(I)$. Since $A^{-1}=\begin{bmatrix}w&-y\\-z&x\end{bmatrix}$, we have 
$$\begin{bmatrix} w&-y\\-z&x\end{bmatrix}\begin{bmatrix} dx &dy\\ dz &dw \end{bmatrix}=\begin{bmatrix} wdx-ydz&wdy-ydw\\-zdx+xdz&\dots\end{bmatrix},$$
so $dx\wedge dy\wedge dz$ pulls back to
\begin{align*}
(wdx-ydz)\wedge (wdy-ydw)\wedge (-zdx+xdz)&=-dx\wedge dz\wedge (w dy-ydw)\\ &=-dx\wedge dz\wedge (w-yz/x)dy \\&=\frac 1x dx\wedge dy\wedge dz.\end{align*}
