Consider $f(x)=x^3$ on the real line, $C=D=\mathbb R$. Then for any smooth $g:E\to \mathbb R$ the pullback $\mathbb R\times_{f,\mathbb R,g}E$ is a smooth manifold diffeomorphic to the graph of $g$, but is is not a submanifold of $\mathbb R\times E$ in general.
So this is not really a counterexample.
Edit:
The following shows, that not every pullback is embedded into the product.
Consider the topological pullback $N = \lbrace(x,y)\in \mathbb R^2: x^2=y^3\rbrace$ of the two smooth mappings $x^2, x^3: \mathbb R\to \mathbb R$ which is Neill's parabola, and consider the manifold $P=\mathbb R$ with the two mappings $x^3, x^2:\mathbb R\to \mathbb R$ which give a topological homeomorphism $P\to N$: $$ \begin{array}{ccccc} P=\mathbb R & \xrightarrow{(x^3,x^2)} & N & \rightarrow & \mathbb R \newline & & \downarrow & & \downarrow x^2 \newline & & \mathbb R & \xrightarrow{x^3} & \mathbb R \end{array} $$ Claim: The triple $(P,x^3,x^2)$ has the universal property of a pullback.
Namely, let $M$ be a smooth manifold and let $f,g:M\to \mathbb R$ be smooth mappings with $f^2= g^3$. Note that then $g\ge 0$. I claim that $f_1:=f^{1/3}:M\to \mathbb R$ is a smooth mapping which gives a smooth factorization $f_1:M\to P$.
Indeed, by convenient calculus (see [1]1) it is sufficient to show, that $f_1\circ c: \mathbb R\to \mathbb R$ is smooth for each smooth curve $c:\mathbb R\to M$. But $(f_1\circ c)^2=g\circ c$ is smooth and $(f_1\circ c)^3 = f\circ c$ is smooth, so by the theorem of Joris (http://mathoverflow.net/questions/127724), $f_1$ is smooth. QED