Why is $\deg(f)$ well-defined? If $M$ and $N$ are boundaryless, compact, connected, oriented $n$-manifolds, and $f:M\to N$ is smooth, then if $\omega_0$ is a $n$-form on $N$ and $\int_N\omega_0\neq 0$, there is a number $a$ such that $\int_M f^*\omega_0=a\int_N\omega_0$.
In fact, if $\omega$ is any $n$-form on $N$, then $\int_M f^*\omega=a\int_N\omega$. How do we know it's the same $a$ for any $\omega$?

Suppose $M$ is oriented, compact and $\partial N=\emptyset$. All $n$-forms on $N$ are closed, and by Stokes theorem, $\int_N\partial \eta=0$ for any exact form, so that the map
$$\int_N:\Omega^n(N)\to\Bbb R$$ automatically factors through the isomorphism
$$\int_N:H^n_{dR}(N)\xrightarrow{\sim\,}\mathbb R\,.$$
The conclusion you are after follows from the fact that the linear map $f^*:H^n_{dR}(N)\to H^n_{dR}(M)$ is multiplication by some real number (actually an integer) $a$.
