The formula in Misner & al (Box 4.1, p. 92, item 4.) is correct, but they could have been clearer in explaining its meaning.
The notation "$\langle\;,\;\rangle$" in Gravitation denotes the dual product (also called "inner product" by some authors) of a vector space and its dual. In this case the vector space is the exterior space $\Lambda^p(V)$ of $p$-vectors based on a vector space $V$ of dimension $n$, and the dual is the exterior space of $p$-covectors $\Lambda^p(V^*)\equiv\Lambda^p(V)^*$.
In coordinate-free form, this dual product is defined as follows:
For a covector $\alpha$ and a vector $a$ it's defined in the usual way;
For a simple $p$-covector (that is, one that can be written as the exterior product of $p$ covectors) $\eta = \eta^1 \wedge \dotsb \wedge \eta^p$, and a simple $p$-vector $w = w_1 \wedge \dotsb \wedge w_p$, it's defined as
$$\langle \eta, w \rangle :=
\eta(w_1, \dotsc, w_p) \equiv
\sum_{\sigma} \mathrm{sgn}(\sigma)\; \langle\eta^{\sigma_1}, w_1\rangle \times \dotsm \times \langle\eta^{\sigma_p}, w_p\rangle,$$
where the sum is over every permutation $\sigma$ of $\{1,\dotsc,p\}$ with sign $\mathrm{sgn}(\sigma)$. It can be shown that this definition doesn't depend on the specific representation of the simple $p$-(co)vector in terms of exterior product of $1$-(co)vectors. Note: simple (co)vectors are also called "monomials" (Choquet-Bruhat & al) or "extensors" (Barnabei & al).
For non-simple $p$-covectors and $p$-vectors it's extended by bilinearity.
From this definition we also obtain the coordinate representation of the dual product. Given a basis $(e_1, \dotsc, e_n)$ for $V$ and dual basis $(\omega^1, \dotsc, \omega^n)$, we build an associated basis $(e_{J_1} \wedge\dotsb\wedge e_{J_p})$ for $\Lambda^p(V)$ and the dual $(\omega^{I_1} \wedge\dotsb\wedge \omega^{I_p})$ for $\Lambda^p(V^*)$. I'm using Choquet-Bruhat & al's convention: capitalized indices $J_k$ mean
$$J_1 < \dotso < J_p$$
(Misner, Thorne, Wheeler use the convention with indices within bars "$\lvert\;\rvert$" instead). This restriction on the indices is very important, because the dimension of $\Lambda^p(V)$ is $\binom{n}{p}$, different from the dimension of the tensor space $\otimes^p V$ in which it could be embedded. Compare the remarks in Choquet-Bruhat & al, p. 197, about "strict components".
These bases are clearly dual from the definition 2. above. Then, by 3., in coordinates the dual product of a $p$-vector with components $A^{J_1\dotso J_p}$ and a $p$-covector $\alpha_{I_1 \dotso I_p}$ is simply
\begin{equation}A^{J_1\dotso J_p}\;\alpha_{J_1 \dotso J_p}\label{dual}\tag{1}
\end{equation}
with Einstein's summation convention.
This is also the expression given in Gravitation.
The crucial step is point 2. above: the dual product does not correspond to a contraction of the antisymmetric tensors corresponding to the $p$-covector and $p$-vector. Formally there's a $p!$ factor difference; but the difference is deeper, because we're operating on different spaces. The exterior spaces $\Lambda^p(V)$ and $\Lambda^p(V^*)$ don't have a tensor product. As Deschamps, § III.4.1, p. 119 says, "The exterior algebra can be introduced independently of tensors and this result is some economy of notation".
Some extra info
The dual product above can be generalized to a product between a $q$-vector $a$ and a $p$-covector $\alpha$; here I denote it with $a \rfloor \alpha$. If $p > q$ the result is a covector, and if $p < q$, a vector. Lindell (§ 1.3.4) calls this the "incomplete dual product"; Deschamps (§ 4.6) the "inner product"; Truesdell & Toupin (§ 267) the "dot product". Its coordinate-free definition for $p > q$ is this: $a \rfloor \alpha$ is the $(p-q)$-covector such that
$$\langle(a\rfloor\alpha),\; x\rangle = \langle \alpha,\; a\wedge x\rangle
\quad\text{for every $(p-q)$-vector $x$}.$$
Its coordinate expression is
$$
(a\rfloor\alpha)_{I_1\dotso I_{p-q}} =
\alpha_{K_1\dotso K_p}\
\epsilon_{J_1\dotso J_q\ I_1\dotso I_{p-q}}^{K_1 \dotso\,\dotso\,\dotso K_p}\
a^{J_1\dotso J_q} \ .
$$
It can be found combining the expression $\eqref{dual}$ for the dual and the expression for the exterior product, given for example by Choquet-Bruhat & al, p. 198 (bottom):
$$(a\wedge x)^{K_1 \dotso K_p} =
\epsilon_{J_1\dotso J_q\ I_1\dotso I_{p-q}}^{K_1 \dotso\,\dotso\,\dotso K_p}\
a^{J_1\dotso J_q} \ x^{I_1\dotso I_{p-q}}.$$
Here $\epsilon$ is the Kronecker tensor (see for example Choquet-Bruhat & al, p. 142).
This product generalizes the dual product discussed above and also the interior product of a $p$-covector $\alpha$ and a $1$-vector $v$, often denoted $\mathrm{i}_v\alpha$. It is often used with the volume element, to transform a $q$-vector into an $(n-q)$-covector (and vice versa with the inverse volume element): see for example Schouten eq. (7.5), § II.7, p. 28. This product also appears, implicitly, in the operation of meet defined in Peano spaces: the meet of two multi-vectors is actually the generalized dual product of both with a volume element; see for example Barnabei & al p. 132, or White p. 98.
This product looks like a contraction, but it has many important differences from the usual contraction in a tensor space. In a tensor space, contraction requires the positional specification of indices or argument slots; in an exterior space this specification doesn't matter (owing to the antisymmetrization).
Finally, the fact that $\langle \omega^{I_1}\wedge\dotsb\wedge\omega^{I_p},
e_{J_1}\wedge\dotsb\wedge e_{J_p}\rangle = \delta^{I_1\dotso I_p}_{J_1\dotso J_p}$– for example $\langle\mathrm{d}x \wedge\mathrm{d}y,
\partial_x \wedge \partial_y\rangle =1$ – makes also sense geometrically: the 2-covector represents a tube with an outer orientation, and the 2-vector a planar area with an inner orientation, and their product tells us how many times the cross-sectional area of the tube fills the planar area (a notion which only requires affine geometry, that is, parallelism). With the dual bases these areas are the same and the result is unity.
The generalized inner product extends this geometric meaning, as beautifully illustrated by Burke (1995, for example p. 17.4) and Schouten (chap. II).
References:
M. Barnabei, A. Brini, G.-C. Rota: On the exterior calculus of invariant theory. J. Algebra 96/1 (1985), 120–160.
W. L. Burke: Div, Grad, Curl Are Dead (1995) http://people.ucsc.edu/~rmont/papers/Burke_DivGradCurl.pdf.
Y. Choquet-Bruhat, C. DeWitt-Morette, M. Dillard-Bleick: Analysis, Manifolds and Physics. Part I: Basics (rev. ed. Elsevier 1996).
G. A. Deschamps, E. M. de Jager, F. John, J. L. Lions, N. Moisseev, F. Sommer, A. N. Tihonov, V. Tikhomirov, A. B. Vasil'eva, V. M. Volosov, D. J. A. Welsh, T. Yamanouchi: Mathematics applied to physics (Springer 1970).
I. V. Lindell: Differential Forms in Electromagnetics (IEEE & Wiley 2004).
C. W. Misner, K. S. Thorne, and J. A. Wheeler: Gravitation (25th printing, Freeman 2003).
J. A. Schouten: Tensor Analysis for Physicists (2nd ed. Dover 1989).
C. A. Truesdell, R. A. Toupin: The Classical Field Theories (Springer 1960, Encyclopedia of Physics Vol. III/1).
N. L. White (ed.): Invariant Methods in Discrete and Computational Geometry (Springer 1995).
Best Answer
First, let's remind ourselves of the action of a tensor product. For any $\omega \in \Lambda^p(V^\star), \theta \in \Lambda^q(V^\star)$, we have
$$ (\omega \otimes \theta)(v_1, \dots, v_p, v_{p+1}, \dots, v_{p+q}) = \omega(v_1, \dots, v_p) \times \theta(v_{p+1}, \dots, v_{p+q}).$$
Let's also recall the definition of the $\text{Alt}$ operator. For any $\lambda \in \Lambda^n(V^\star)$, we have
$$ \text{Alt}(\lambda)(v_1, \dots, v_n) = \frac{1}{n!}\sum_{\sigma \in S_n}\text{sgn}(\sigma)\lambda(v_{\sigma(1)}, \dots, v_{\sigma(n)}). $$
The wedge product is constructed by applying the $\text{Alt}$ operator to the tensor product, i.e. $$ \omega \wedge\theta = \frac{(p+q)!}{p!q!}\text{Alt}(\omega \otimes \theta).$$
So $$ (\omega \wedge \theta)(v_1, \dots, v_p, v_{p+1}, \dots, v_{p+q}) = \frac{1}{p!q!} \sum_{\sigma \in S_{p+q}} \text{sgn} (\sigma) \omega(v_{\sigma(1)}, \dots, v_{\sigma(p)}) \theta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}). $$
Thus $f^\star(\omega \wedge \theta)(v_1, \dots, v_p, v_{p+1}, \dots, v_{p+q}) = (f^\star\omega \wedge f^\star\theta)(v_1, \dots, v_p, v_{p+1}, \dots, v_{p+q})$, since both of these quantities are equal to $$ \frac{1}{p!q!} \sum_{\sigma \in S_{p+q}} \text{sgn} (\sigma) \omega(f_\star (v_{\sigma(1)}), \dots, f_\star(v_{\sigma(p)})) \theta(f_\star(v_{\sigma(p+1)}), \dots, f_\star(v_{\sigma(p+q)})). $$
Edit: I notice you've added an extra question at the bottom of your post about $dx \wedge dy \left(\frac{\partial}{\partial y}, \frac{\partial}{\partial x} \right)$. This is a great example for illustration, so let's work through it. The calculation is:
\begin{align*} dx \wedge dy \left(\tfrac{\partial}{\partial y}, \tfrac{\partial}{\partial x} \right) & = \frac{1}{1!\times 1!} \times \left( dx \left(\tfrac{\partial}{\partial y}\right) dy \left( \tfrac{\partial}{\partial x} \right) - dx \left(\tfrac{\partial}{\partial x}\right) dy \left(\tfrac{\partial}{\partial y} \right) \right) \\ & = 1 \times (0 - 1) \\ & = -1 \end{align*}