Calculating a Multivariable derivative. I'm trying to work through Spivak's Calculus on Manifolds and I've arrived at Differentiation.  While I can usually follow his steps, I find myself lost or stuck when I try to do something on my own.  So I decided to work through one of his first examples using $Df$ notation instead of $f'$ notation.
My main point, I have confused myself.  My question is clearly asked only at the very bottom of this post.
As for the example, I need to calculate the derivative of $f:\mathbb{R}^{2}\to \mathbb{R}$,
where
$$f(x,y) = \sin(xy^2).$$
The following rules are available to me:
1) For a point $a$ in the domain of $f$ such that $f(a)$ is in the domain of $g$,
$$D(g\circ f)(a) = Dg(f(a))\circ Df(a).$$
2) For two functions $f,g:\mathbb{R}^{n}\to \mathbb{R}$, 
$$D(fg)(a) = g(a)Df(a) + f(a)Dg(a)$$ and $$D(f+g)(a) = Df(a) + Dg(a).$$
If I have stated either of these rules even slightly incorrectly please be brutally in my face about it.
I'm trying to carefully apply this rules to my function.
If I let $p,s:\mathbb{R}^{2}\to \mathbb{R}$ denote the product function and $s:\mathbb{R}\to \mathbb{R}$ represent the squaring function, I can write:
$f = \sin\circ p\circ (\pi_{1}, s\circ \pi_{2})$, where $\pi_{1}$ and $\pi_{2}$ are the coordinate functions.
Now my derivative of $f$, denoted $Df$, should be a map from $\mathbb{R}^{2}\to \mathbb{R}$, just like $f$ is.
So at a point $(a,b)\in \mathbb{R}^{2}$, I can write
\begin{align*}
Df(a,b) &= D\left(\sin\circ p\circ (\pi_{1}, s\circ \pi_{2})\right)(a,b)\\
        &= D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b))\circ Dp((\pi_{1}, s\circ \pi_{2})(a,b))\circ D(\pi_{1}, s\circ \pi_{2})(a,b)
\end{align*}
So I try to calculate this in separate blocks:
\begin{align*}
D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b)) &=  \cos(p\circ (\pi_{1}, s\circ \pi_{2})(a,b))\\
&= \cos(p\circ (\pi_{1}(a,b), [s\circ \pi_{2}](a,b)))\\
&= \cos(p\circ (a, s(b)))\\
&= \cos(p\circ (a, b^2)))\\
&= \cos(ab^2).
\end{align*}
But this brings me to my first (among several) points of confusion.
In the equation:
$$Df(a,b) = D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b))\circ Dp((\pi_{1}, s\circ \pi_{2})(a,b))\circ D(\pi_{1}, s\circ \pi_{2})(a,b)$$
it appears $D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b))$ should be a function, not a number.  Can someone point out what my error in thinking is? (answered below)
Continuing on to compute the 3rd block, 
\begin{align*}
D(\pi_{1}, s\circ \pi_{2})(a,b) &= (D\pi_{1}(a,b), D(s\circ \pi_{2})(a,b))\\
&= (\pi_{1}(a,b), Ds(\pi_{2}(a,b))\circ D\pi_{2}(a,b))\\
&= (a, Ds(b)\circ \pi_{2}(a,b))\\
&= (a, 2b\circ b)\\
&= (a, 2b^2)
\end{align*}
Now the middle one:
\begin{align*}
Dp((\pi_{1}, s\circ\pi_{2})(a,b)) &= Dp((\pi_{1}(a,b), (s\circ \pi_{2})(a,b))\\
&= Dp(a, b^2)
\end{align*}
Now substituting these smaller calculations, the whole thing simplifies down to:
\begin{align*}
Df(a,b) &= D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b))\circ Dp((\pi_{1}, s\circ \pi_{2})(a,b))\circ D(\pi_{1}, s\circ \pi_{2})(a,b)\\
&= \cos(ab^2)\circ \underbrace{Dp(a, b^2)\circ (a, 2b^2)}_{= a\cdot 2b^2 + b^2\cdot a}\\
&= \cos(ab^2)(3ab^2)
\end{align*}
Now I will insist that I have something wrong.  $Df(a,b)$ should be a map from $\mathbb{R}^{2}\to \mathbb{R}$. But it has collapsed into a single real number.
Spivak calculates the derivative using Jacobian notation, arriving at the conclusion that
$f'(a,b) = (b^2\cdot\cos(ab^2), 2ab\cdot \cos(ab^2))$, which naturally is the transformation matrix for a map $\mathbb{R}^{2}\to \mathbb{R}$.
Sorry this problem is so long winded, but I wanted to show all my steps so as to be able to identify the one that went awry.
 A: This is a typo in the book. You can see in the Remark after the chain rule theorem $2$-$2$ that that composition symbol was supposed to be a multiplication dot.
A: If $f: \mathbb{R}^m \longrightarrow \mathbb{R}^n$ is differentiable at $p \in \mathbb{R}^n$, then its derivative $Df(p)$ is the linear map
$$Df(p) : \mathbb{R}^n \longrightarrow \mathbb{R}^m$$
such that
$$\lim_{h \to 0} \frac{\|f(p + h) - f(p) - Df(p)h\|}{\|h\|} = 0.$$
For what you are asking, $\sin: \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable everywhere, so at $p\circ (\pi_{1}, s\circ \pi_{2})(a,b) = ab^2 \in \mathbb{R}$, its derivative is a linear map
$$D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b)): \mathbb{R} \longrightarrow \mathbb{R}.$$
Linear maps from $\mathbb{R}$ to $\mathbb{R}$ are just multiplication by scalars. In this case, for any vector $v \in \mathbb{R}$, we have
$$D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b))(v) = \cos(ab^2)v.$$
This is how $D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b))$ is defined as a function from $\mathbb{R}$ to $\mathbb{R}$.

With regards to your edit:
$(\pi_1, s \circ \pi_2)$ is a map from $\mathbb{R}^2$ to $\mathbb{R}^2$:
$$(\pi_1, s \circ \pi_2)(x,y) = (x, y^2).$$
Then $D(\pi_1, s \circ \pi_2)(a,b)$ should be a linear map from $\mathbb{R}^2$ to $\mathbb{R}^2$. You obtained $(a, 2b^2)$ for this derivative, which is a map from $\mathbb{R}^2$ to $\mathbb{R}$. You should instead find
$$D(\pi_1, s \circ \pi_2)(a,b) = \begin{pmatrix} 1 & 0 \\ 0 & 2b \end{pmatrix}.$$
You didn't write out what $Dp$ is either. You should find that
$$Dp(a,b) = (b,a).$$
Putting this all together, you have that
\begin{align*}
Df(a,b) & = D(\sin)(p\circ (\pi_{1}, s\circ \pi_{2})(a,b)) \circ Dp((\pi_{1}, s\circ \pi_{2})(a,b)) \circ D(\pi_1, s \circ \pi_2)(a,b) \\
 & = D(\sin)(ab^2) \circ Dp(a,b^2) \circ D(\pi_1, s \circ \pi_2)(a,b) \\
 & = \cos(ab^2) (b^2, a) \begin{pmatrix} 1 & 0 \\ 0 & 2b \end{pmatrix} \\
 & = \cos(ab^2) (b^2, 2ab) \\
 & = (b^2 \cos(ab^2), 2ab \cos(ab^2)),
\end{align*}
as desired.
A: Sorry to have wasted anyone's time.  My error was the following, misquoting a previous theorem about $Df(a)$ for a linear transformation $f$:
When I say $D\pi_{1}(a,b) = \pi_{1}(a,b)$, this is false.  The corrected version of the statement is $D\pi_{1}(a,b) = \pi_{1}$.  This fixes the problem I think.
