After further study of Spivak and other texts, I take back my original comment that there is a typo. This is an example where there is technically no error, but poor notation and Spivak's famous terseness makes the book extremely confusing for students.
On page 77, Spivak defines the pullback of a tensor $T\in\mathcal{T}^k(W)$ by a linear transformation $f:V\to W$ as $f^*T(v_1,\ldots,v_k)=T(f(v_1),\ldots, f(v_k))$.
On page 89, he defines $f_*:\mathbb{R}^n_p\to \mathbb{R}^m_{f(p)}$ to be the differential of smooth map $f:\mathbb{R}^n\to\mathbb{R}^m$ as $f_*(v_p)=(Df(p)(v))_{f(p)}$. Given this definition, we can pull back the $\it{value}$ of a differential form $\omega$ at $f(p)$, $\omega(f(p))\in\mathcal{A}^k(\mathbb{R}^m_{f(p)})$ (which is an alternating tensor on $\mathbb{R}^m_{f(p)}$), by $f_*$, by defining $(f_*)^*(\omega(f(p)))$ for $v_{1p},\ldots, v_{kp}
\in \mathbb{R}^n_p$ as
$$[(f_*)^* (\omega(f(p)))](v_{1p},\ldots,v_{kp})=
[\omega(f(p))](f_*(v_{1p}),\ldots,f_*(v_{kp})),$$
as an application of the definition on p. 77.
Note that $(f_*)^* (\omega(f(p)))\in\mathcal{A}^k(\mathbb{R}^n_p)$. Thus, $(f_*)^*$ is a map $\mathcal{A}^k(\mathbb{R}^m_{f(p)})\to \mathcal{A}^k(\mathbb{R}^n_p)$. This is the induced linear transformation $f^*$ that Spivak talks about.
What Spivak fails to mention is that $f^*\equiv (f_*)^*$ is the conventional simplified notation of this map $(f_*)^*$. The confusion comes from the fact that the reader was introduced to the related but different definition of $f^*$ on p. 77. Baffling to the novice though accurate, Spivak introduces $f^*$ as the map induced by $f_*$ without mentioning its relation to the p. 77 definition or this notational convention.
If this is still confusing, the preceding discussion can be ignored: the definition of $f^*$ useful in practice is the one that he gives at the top of p. 90:
$$f^*\omega(p)(v_{1p},\ldots, v_{kp})=\omega(f(p))(f_*(v_{1p}),\ldots,
f_*(v_{kp})).$$
Finally, what Spivak means by $(f^*\omega)(p)=f^*(\omega(f(p)))$ is that he is defining a new differential form, $f^*\omega$ by $\it{pointwise}$ pullback of each tensor $\omega(f(p))$. Just to be clear, for each point $p$, $(f^*\omega)(p)\in\mathcal{A}^k(\mathbb{R}^n_p)$, so that $f^*\omega$ is a $k$-form on $\mathbb{R}^n$, while $\omega$ is a $k$-form on $\mathbb{R}^m$.
For detailed discussions of definitions, see: Volume 1 of A Comprehensive Introduction to Differential Geometry by Spivak (where all the confusing things in Calculus on Manifolds are actually explained by the author himself) or An Introduction to Manifolds by Loring Tu, which is probably the clearest introduction to smooth manifold theory for first-year graduate students and advanced undergrads.
PREVIOUS REPLY:
There's a typo on p. 89: the definition of the pullback of a differential form $\omega$ should be $(f^*\omega)(p)=(f_*)^* (\omega(f(p)))$. The rhs of this definition is based on the definition of the pullback of a tensor, given on p. 77: for a map $f:V\to W$ and a tensor $T\in \mathcal{T}^k(W)$, the pullback of $T$ is
$f^*T\in\mathcal{T}^k(V)$ defined by $$(f^*T)(v_1,\ldots, v_k)=T(f(v_1),
\ldots, f(v_k)).$$
Unpacking these definitions, if $v_{1p},\ldots,v_{kp}\in\mathbb{R}^n_p$,
we have $$[(f^*\omega)(p)](v_{1p},\ldots,v_{kp})=[(f_*)^* (\omega(f(p)))](v_{1p},\ldots,v_{kp})=
[\omega(f(p))](f_*(v_{1p}),\ldots,f_*(v_{kp})),$$
as claimed by Spivak.
Best Answer
You should note that $f'$ (in your notation) is the total derivative of the differentiable map $f:U\to V$. Typically, this is written $df$ or $Df$. Indeed, at a point $p\in U$, $df_p:T_pU\to T_pV$ transforms tangent vectors to tangent vectors. In local coordinates, $df_p$ is the Jacobian matrix of the transformation - i.e. the matrix of first partials.
It's good that you tried an example, and you are somewhat close. The only thing you are missing is how the total differential $df$ acts. Indeed, it acts as a linear transformation on elements of the tangent space. So, let's take $v\in T_t\mathbb{R}$ in your example. Because $v$ is in the tangent space to $\mathbb{R}$, this is of course just a scalar, i.e. a real number. Then, as you correctly calculated: $$ df_t = ((\sin{t} + t\cos{t})e^{t\sin{t}}, 2t - 2\pi, -\frac{1}{2}\sin{\frac{t}{2}}).$$ We apply this to $v$ to get $df_t(v)= ((\sin{t} + t\cos{t})e^{t\sin{t}}v, (2t - 2\pi)v , -\frac{1}{2}\sin({\frac{t}{2}})v)$ i.e. $$ df_t(v)=((\sin t+t\cos t)e^{t\sin t}v)\frac{\partial}{\partial x}+((2t-2\pi)v)\frac{\partial}{\partial y}-(\frac{1}{2}\sin (\frac{t}{2})v)\frac{\partial}{\partial z}.$$ Then, \begin{align*} (f^{*}\omega)_t(v) &= \left(e^{t\sin{t}}dx + (t^2 - 2\pi t)dy + \cos{\frac{t}{2}}dz\right)df_t(v)\\ &=\left(e^{t\sin{t}}dx + (t^2 - 2\pi t)dy + \cos{\frac{t}{2}}dz\right)\\ &\left(((\sin t+t\cos t)e^{t\sin t}v)\frac{\partial}{\partial x}+((2t-2\pi)v)\frac{\partial}{\partial y}-(\frac{1}{2}\sin (\frac{t}{2})v)\frac{\partial}{\partial z}\right)\\ &=(\sin t+t\cos t)e^{2t\sin t}v+(2t-2\pi)(t^2-2\pi t)v-\frac{1}{2}\cos(\frac{t}{2})\sin(\frac{t}{2})v. \end{align*} Note that this unsightly expression is in fact a scalar. This is good, because our differential form evaluated at $t$ is a linear functional $(f^*\omega)_t\in (T_t\mathbb{R})^*$.
The moral of the story is the following: when $f:U\to V$ is given as above, to evaluate $(f^*\omega)_p$ on $v\in T_pU$, you push $v$ forward with $df_p$ then evaluate $\omega$ on it: $$ (f^*\omega)_p(v)=\omega_{f(p)}(df_pv).$$ I apologize for the unsightly formulas. If something seems wrong - computationally speaking or otherwise - please let me know.