$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\renewcommand{\phi}{\varphi}$As you may know, differential forms on $M$ can be "vector-valued", i.e., they can take values in some vector bundle $E \to M$. In the simplest case, $E$ is the trivial real line bundle, and "$E$-valued differential forms" are just "differential forms". In general, if $(\Basis_{j})_{j=1}^{\ell}$ is a local frame for $E$ in some trivializing coordinate neighborhood $U$ (i.e., both $E$ and $TM$ are trivial over $U$), an $E$-valued $k$-form looks like
$$
\sum_{j=1}^{\ell} \omega_{j} \Basis_{j}
$$
with the $\omega_{j}$ ordinary $k$-forms in $U$.[1]
If $\phi:M \to N$ is a smooth map, and if $p:E \to N$ is a vector bundle, there is a pullback bundle $\phi^{*}E \to M$ whose total space is
$$
\{(x, v) \text{ in } M \times E: \phi(x) = p(v)\}
$$
and whose projection map is projection to the first factor. Intuitively, put a copy of the fibre $E_{\phi(x)}$ over $x$ for each $x$ in $M$.
The push-forward is sometimes introduced informally as a mapping $\phi_{*}:TM \to TN$, but technically that's not right. Actually, $\phi_{*}$ takes values in $\phi^{*}TN$ and is a mapping between vector bundles over $M$.
The differential $d\phi$ may be viewed as a $1$-form on $M$ taking values in $\phi^{*}TN$, the pullback of the tangent bundle by $\phi$. That's a fancy way of saying that if $v$ is a tangent vector to $M$ at a point $x$, then $d\phi(x)(v)$ is an element of $T_{\phi(x)}N$, and is linear in $v$. (Of course, $d\phi(x)(v)$ actually measures something useful, the "rate of change" of $\phi$ at $x$ in the direction $v$.)
For a real-valued function $f$, the differential $df$ is indeed the exterior derivative, by the usual definition of the exterior derivative. To mesh this with the preceding item, note that $T\Reals$ is a trivialized vector bundle: We know what "$1$" means when we speak of real-valued functions and forms, so $T\Reals = \Reals \times \Reals$ and consequently $f^{*}T\Reals = M \times \Reals$. That means we can view the differential $df$ as a $1$-form with values in $M \times \Reals$, i.e., as a real-valued $1$-form.
Your third bullet point may be answered by the first bullet point above, but it's probably worth mentioning that a smooth map $\phi:M \to N$ could be termed an "$N$-valued $0$-form", whose differential is $d\phi$.
Notationally, I think of $d\phi(x)$ as the linear fibre map sending a tangent vector $v$ in $T_{x}M$ to $d\phi(x)(v)$ in $T_{\phi(x)}N$, while the push-forward $\phi_{*}$ is the bundle map sending $(x, v)$ in $TM$ to $\bigl(\phi(x), d\phi(x)(v)\bigr)$ in $TN$. So, they're not quite identical if you need to split hairs, but in practice you can usually talk only about the push-forward.
- Technically, an $E$-valued differential $k$-form is a smooth section of the vector bundle $\bigwedge^{k}T^{*}M \otimes E$. If you change trivializations, the form "coefficients" transform like ordinary differential forms, and the "vector parts" transform like sections of $E$. In general, there's no natural exterior derivative on $E$-valued forms; you have to know how to differentiate sections of $E$, as well. If $E$ has locally constant transition functions, however, then there is a natural exterior derivative, given by
$$
d\left(\sum \omega_{j} \Basis_{j}\right) = \sum (d\omega_{j}) \Basis_{j}.
$$
One can't mention this without making a plug for holomorphic vector bundles: Holomorphic functions act like constants with respect to the Dolbeault operator $\bar{\partial}$ (i.e., $f$ is holomorphic if and only if $\bar{\partial}f = 0$), so there's a well-defined $\bar{\partial}$ operator
$$
\bar{\partial}\left(\sum \omega_{j} \Basis_{j}\right) = \sum (\bar{\partial}\omega_{j}) \Basis_{j}
$$
for differential forms with values in a holomorphic vector bundle $E$. (!)
