Steve has more or less already answered to your question.

Just to give you some more intuition, let $M$ be an oriented smooth real manifold of real dimension $m$ and consider the space $^s\mathcal E^p(M)$, $s\in\mathbb N\cup\{+\infty\}$, of differential forms $C^s(M,\Lambda^p T^*_M)$ endowed with the topology of uniform convergence of all the derivatives of order $\le s$ of the coefficients on every compact subset (contained in a coordinate chart) of $M$. If $K\subset M$ is a compact subset, let $^s\mathcal D^p(K)$ denote the subspace of elements $u\in ^s\mathcal E^p(M)$ with support contained in $K$, together with the induced topology. Finally, put
$$
^s\mathcal D^p(M)=\bigcup_{K\subset\subset M} ^s\mathcal D^p(K).
$$
The space $^s\mathcal D'_p(M)=^s\mathcal D'^{m-p}(M)$ is by definition the topological dual $(^s\mathcal D_p(M))'$ and is called the space of *currents* of dimension $p$ (or degree $m-p$) and order $s$ on $M$.

You should think of currents as a generalization of differential forms (locally they are differential forms with distributional coefficients), precisely how distributions generalize the concept of functions.

**Example 1.** If $f$ is a differential form of degree $q$ with $L^1_{\text{loc}}$ coefficients, we can associate to $f$ the current $T_f$ of dimension $m-q$
$$
\langle T_f,u\rangle=\int_Mf\wedge u,\quad u\in ^0\mathcal D^{m-q}(M).
$$
$T_f$ is of degree $q$ and of order $0$. The correspondence $f\mapsto T_f$ is injective.

**Example 2.** Let $Z\subset M$ be a closed oriented submanifold of $M$ of dimension $p$ and class $C^1$ (here $Z$ may have a boundary $\partial Z$). The current of integration over $Z$, denoted $[Z]$, is defined by
$$
\langle[Z],u\rangle=\int_Z u,\quad u\in ^0\mathcal D^p(M).
$$
It is clear that $[Z]$ is a current of order $0$ on $M$ and that $\operatorname{Supp}(Z)=Z$ (the support of a current is the smallest closed subset $A\subset M$ such that the restriction of the current to forms with compact support contained in the complementary of $A$ is zero). Its dimension is $p=\dim Z$.

The exterior derivative $dT\in ^{s+1}\mathcal D'^{q+1}(M)=^{s+1}\mathcal D'_{m-q-1}$ of a current $T\in ^s\mathcal D'^q(M)=^s\mathcal D'_{m-q}$ is defined by
$$
\langle dT,u\rangle=(-1)^{q+1}\langle T,du\rangle,\quad u\in ^{s+1}\mathcal D^{m-q-1}(M).
$$
For all forms $f\in ^1\mathcal E^q(M)$ and $u\in ^1\mathcal D^{m-q-1}(M)$, Stokes' formula implies
$$
0=\int_M d(f\wedge u)=\int_M df\wedge u+(-1)^qf\wedge du,
$$
thus in our Example 1 we actually have $dT_f=T_{df}$ as it should be.

Finally, and this should clarify your question, in our Example $2$ another application of Stokes' formula gives
$$
\int_Z du=\int_{\partial Z}u,
$$
therefore $\langle[Z],du\rangle=\langle[\partial Z],u\rangle$ and $d[Z]=(-1)^{m-p+1}[\partial Z]$.