In this problem, it is assumed that $M$ is orientable. My question is: why is this needed? Is this a mistake in the book, or is my solution wrong somehow?
18-1 (paraphrased): Suppose that $M$ is an oriented smooth manifold and $\omega$ is a closed $p$-form on $M$.
(a) Show that $\omega$ is exact if and only if the integral of $\omega$ over every smooth $p$-cycle is zero.
(b) Suppose that the $p$-th singular homology $H_p(M)$ is finitely generated by the smooth $p$-cycles $\{c_1,\ldots,c_m\}$. Show that $\omega$ is exact if and only if for each $i$, $\int_{c_i}\omega = 0$.
My solution:
(a) If $\omega = d\eta$ is exact, then by Stokes's theorem for chains, $\int_c\omega = \int_c d\eta = \int_{\partial c}\eta = \int_0 \eta = 0$, since $\partial c = 0$ for any $p$-cycle $c$. Conversely if $\int_c \omega = 0$ for any cycle $c$, then $c \mapsto \int_c\omega$ represents $0$ in singular cohomology. By de Rham's theorem, $\omega$ therefore is zero in de Rham cohomology, so $\omega$ is exact.
(b) If $\omega$ is exact, then the integral of $\omega$ over any $c_i$ is zero by the same proof in part (a). Conversely if for all $i$: $\int_{c_i}\omega = 0$, then by linearity $\int_c \omega = 0$ for any $p$-cycle $c$, since $c$ can be written as a linear combination of the generators $\{c_1,\ldots,c_m\}$. Hence by the proof in part (a), $\omega$ is exact. $\square$
Remark: Theorem 18.12 ("Stokes's Theorem for Chains") on p.481 of Lee's book does not assume orientability of $M$.
You're right -- there's no need for the "oriented" hypothesis. Good catch. I've added this to my correction list.