To really connect the claims I make below with the definitions given in your post takes some effort, but since you asked for intuition here it goes...
I think the best intuition for push-forwards and pull-backs is offered from the case $\mathcal{M}=\mathcal{N} $ and both are embedded in $\mathbb{R}^n$.
the push-forward amounts to changing coordinates for a vector field (viewed as a derivation) for example: for cartesians verses polars in $\mathbb{R}^2$
$$ \frac{\partial}{\partial r} = \frac{\partial x}{\partial r}\frac{\partial}{\partial x}+
\frac{\partial y}{\partial r}\frac{\partial}{\partial y} = \cos(\theta)\frac{\partial}{\partial x}+\sin(\theta)\frac{\partial}{\partial y}$$
the pull-back amounts to taking the total differential of the coordinate formulas, or equivalently, applying the chain-rule to swap differentials in one coordinate system for another. For example, again for two-dimensional Cartesian verse polars:
$$ dr = d(\sqrt{x^2+y^2}) =\frac{x}{\sqrt{x^2+y^2} }dx+\frac{y}{\sqrt{x^2+y^2} }dy $$
The role of the pullback to integration is that it allows us to lift integration defined in $\mathbb{R}^n$ up to the manifold (provided we have the partition of unity to weave things together). Essentially the abstract $dx$ on the manifold pulls down to an ordinary differential on $\mathbb{R}^n$. However, it's not quite that simple because we have $dx \wedge dy = -dy \wedge dx$ on the manifold. In an integral on $\mathbb{R}$ we have $dxdy=dydx$ provided we swap the integration bounds appropriately. This apparent contradiction is reconciled by the fact that integrals of two-forms correspond to surface integrals. In a surface integral the order of the parameters determines the orientation of the surface. All things being otherwise the same $f(x,y)dx \wedge dy$ and $f(x,y)dy \wedge dx$ pull back to the same surface integral modulo a sign. Of course this sign is important, it measures outward or inward flux for the example I'm currently discussing.
All of this said, it bugs me to no end when texts choose to omit the $\wedge$ for explicit form calculations!
Per request: an explicit example
Suppose we take $\phi (A) = (A_{11},A_{12},A_{21},A_{22})$ as the coordinate chart on $\mathcal{M} = \mathbb{R}^{ 2 \times 2}$. Denote the coordinate functions by $x_{ij}: \mathcal{M} \rightarrow \mathbb{R}^4$. To integrate in this 4-dimensional manifold we need to find some 4-form with which to play. Define,
$$ \omega = (x_{11}^2+2)dx_{11} \wedge dx_{12} \wedge dx_{21} \wedge dx_{22} $$
To help separate concepts let me use $(u_1,u_2,u_3,u_4)$ as the standard cartesian coordinates on $\mathbb{R}^4$. Let's calculate the pull-back of $\omega$ under $\phi^{-1}=\Psi$. Since $\Psi^*(\omega)$ will be a four-form at $p=(p_1,p_2,p_3,p_4)$ on $\mathbb{R}^4$ it suffices to consider (suppressing the $p$ for now)
$$ \Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(\partial_1),
\Psi_*(\partial_2),\Psi_*(\partial_2),\Psi_*(\partial_4))$$
We need to calculate the push-forwards to continue,
$$ \Psi_*(\partial_1) = \partial_{11}, \ \
\Psi_*(\partial_2) = \partial_{12}, \ \
\Psi_*(\partial_3) = \partial_{21}, \ \
\Psi_*(\partial_4) = \partial_{22} $$
Thus,
\begin{align} \Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4) &=\omega (\partial_{11},\partial_{12},\partial_{21},\partial_{22}) \\
&= (x_{11}^2+2)(dx_{11} \wedge dx_{12} \wedge dx_{21} \wedge dx_{22}) (\partial_{11},\partial_{12},\partial_{21},\partial_{22}) \\
&= x_{11}^2+2
\end{align}
Following the $p$ through the calculation will reveal that we find:
$$ \Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)_p = p_1^2+2 $$
Or, if you like, set $p=u$ and find $u_1^2+2$. To perform an integration we best take some compact subset of the matrix space then you can see that it will simply reduce to integrating the function $u_1^2+2$ on the corresponding parameter space in $\mathbb{R}^4$
About half way through this I realized you only wanted a covector example. For that, we simply need an abstract one-dimensional manifold and a one-form with which to play. If you wish I'll add that later.
Instead of thinking of $\alpha$ as a map, think of it as a substitution of variables:
$$
x = uv,\qquad y=u^2,\qquad z =3u+v.
$$
Then
$$
dx \;=\; \frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv \;=\; v\,du+u\,dv
$$
and similarly
$$
dy \;=\; 2u\,du\qquad\text{and}\qquad dz\;=\;3\,du+dv.
$$
Therefore,
$$
\begin{align*}
xy\,dx + 2z\,dy - y\,dz \;&=\; (uv)(u^2)(v\,du+u\,dv)+2(3u+v)(2u\,du)-(u^2)(3\,du+dv)\\[1ex]
&=\; (u^3v^2+9u^2+4uv)\,du\,+\,(u^4v-u^2)\,dv.
\end{align*}
$$
We conclude that
$$
\alpha^*(xy\,dx + 2z\,dy - y\,dz) \;=\; (u^3v^2+9u^2+4uv)\,du\,+\,(u^4v-u^2)\,dv.
$$
Best Answer
Suppose you have a map $\alpha: A \rightarrow B$. There are certain associated items that "go forward" or "go backwards":
Points are sent forward. Given $p \in A$ we have $\alpha(p) \in B$.
Functions are sent back, i.e. pull back from $B$ to $A$. If we have a function $f: B \rightarrow \mathbb{R}$ then we get the composition $f \circ \alpha : A \rightarrow \mathbb{R}$.
Vectors are send forward. Given $v \in TA_p$ we have the derivative map $d\alpha_p: TA_p \rightarrow TB_{\alpha(p)}$ (or maybe your book calls this $\alpha_*$; in coordinates it's just the derivative matrix). So we get $d\alpha_p(v) \in TB_{\alpha(p)}$.
One-forms give linear functionals on each tangent space; that is, they're functions which take vectors as input. As such, just like functions, they're sent back. If we have a one-form $\omega$ on $B$, then we get $\omega_q : TB_q \rightarrow \mathbb{R}$. So given $p \in A$ with $\alpha(p) = q$ we get $\omega_q \circ d\alpha_p : TA_p \rightarrow \mathbb{R}$ is a linear functional on $TA_p$. We get a one-form on $A$ this way.
Basically, geometric objects "go forward" and functions on them "go back."