This may be a situation where it helps to think in coordinates. First, I establish these coordinates.
Fix an atlas of charts $(U_{\alpha},\phi_{\alpha})_{\alpha \in A}$ on $M$. We can assume each $\phi_{\alpha}$ identifies $U_{\alpha}$ with $\mathbf{R}^{m}$. An atlas of charts on the compact unit interval $[0,1]$ is given by $t \mapsto t$ on $[0,1)$ and $t \mapsto 1 - t$ on $(0,1]$. Take the product to obtain charts $(U_{\alpha,i},\phi_{\alpha,i})_{\alpha \in A, i \in \{0,1\}}$ on the product manifold $M \times [0,1]$.
Explicitly, $U_{\alpha,0} = U_{\alpha} \times [0,1)$, and the map $\phi_{\alpha,0}\colon U_{\alpha,0} \to \mathbf{R}^{m} \times [0,1)$ maps $(p,t)$ to $(\phi_{\alpha}(p),t)$. Likewise, $U_{\alpha,1} = U_{\alpha} \times (0,1]$, and the map $\phi_{\alpha,1}\colon U_{\alpha,1} \to \mathbf{R}^{m} \times [0,1)$ maps $(p,t)$ to $(\phi_{\alpha}(p),1-t)$.
Notice that, for a vector $v \in \operatorname{T}_{(p,t)}(M \times [0,1])$, i.e., a vector tangent to $M \times [0,1]$ at the point $(p,t)$, the identifications $\phi_{\alpha,0,*}(v)$ and $\phi_{\alpha,1,*}(v)$ with vectors tangent to $\mathbf{R}^{n} \times [0,1)$ at $(\phi_{\alpha}(p),t)$ and $(\phi_{\alpha}(p),1-t)$, respectively, differ in sign in the final coordinate. Specifically, if $\phi_{\alpha,0,*}(v) = \partial_{t}$, then $\phi_{\alpha,1,*}(v) = -\partial_{t}$. In matrix form, the tangent spaces to $\mathbf{R}^{n} \times [0,1)$ are related by
$$ \begin{pmatrix} I_{m} & 0 \\ 0 & -1 \end{pmatrix} $$
with respect to the canonical basis $\partial_{1},\dots,\partial_{m},\partial_{t}$.
The key point is that $M \times \{1\}$ as an oriented manifold is not $M \times \{0\}$, but rather $M \times \{0\}$ with the opposite orientation. To see this, consider the oriented trivialization $\partial_{1},\dots,\partial_{n},\partial_{t}$ of the tangent bundle to $\mathbf{R}^{n} \times [0,1)$. If this trivialization pulls back along $\phi_{\alpha,0}$ to an oriented trivialization of the tangent bundle to $U_{\alpha,0} \times [0,1)$, i.e., $\phi_{\alpha,0}$ is orientation-preserving, then the same trivialization pulled back instead along $\phi_{\alpha,1}$ reveals that the same orientation is given by $\partial_{1},\dots,\partial_{n},-\partial_{t}$ on the chart $(U_{\alpha,1},\phi_{\alpha,1})$. This in turn induces the following orientation on $(M \times \{1\}) \cap U_{\alpha,1}$.
Let $v_{1},\dots,v_{m}$ be a basis of vectors tangent to $M \times \{1\}$ at a point $(p,1)$ for some $p \in U_{\alpha}$. This basis is oriented with respect to the orientation above if, after pushing forward along $\phi_{\alpha,1}$, we obtain vectors $(\phi_{\alpha,1,*}v_{1}),\dots,(\phi_{\alpha,1,*}v_{m}),\partial_{t}$ conjugate to $\partial_{1},\dots,\partial_{m},-\partial_{t}$ by some element of $\operatorname{GL}_{m}^{+}(\mathbf{R})$, i.e., a matrix of positive determinant. Note that this is not the orientation induced by the chart $\phi_{\alpha,1}$. The latter would instead demand conjugacy to the basis $\partial_{1},\dots,\partial_{m},\partial_{t}$.
We can think of the former orientation as that obtained by translating the orientation from one end of the cylinder to the other, and the latter orientation as that determined by the inward-pointing normal. If we identify $M \times \{0\}$ with $M \times \{1\}$ by translation, i.e., by projecting onto the first factor, we are imposing the former on $M \times \{1\}$. A consequence of this is that the inclusion $i_{1}$ is then orientation-reversing. A subsequent consequence is that, if we integrate by pulling back along $i_{1}$, then we need to compensate for the reversal of orientation with a negative sign. This is the change-of-variables formula.
Finally, I turn to the $m$-form $h^{*}\omega$ on $M \times [0,1]$. In fact, the statement has little to do with $h$ and more to do with the inclusions $i_{0}$ and $i_{1}$ of the boundary components, so I take $N = M \times [0,1]$ and $h$ to be the identity for the sake of simplicity. Indeed, I assert that for any $m$-form $\omega$ on $M \times [0,1]$,
$$ \int_{M} i_{1}^{*}\omega - \int_{M} i_{0}^{*}\omega = \int_{\partial(M \times [0,1])} \omega. $$
Begin from the right hand side. If we identify $M$ with $M \times \{1\}$, then
\begin{align*}
\int_{\partial(M \times [0,1])} \omega &= \int_{M \times \{0\}} \omega + \int_{M \times \{1\}} \omega \\
&= -\int_{M} i_{0}^{*}\omega + \int_{M} i_{1}^{*}\omega.
\end{align*}
Thus, the equation appears.
As a final note, this observation and Stokes's theorem together yield the fundamental theorem of calculus if one takes $M$ to be the one-point space and $N$ to be $\mathbf{R}$.
$$ \int_{[0,1]} f'(x)dx = \int_{\partial([0,1])} f(x) = \int_{\{1\}} f(x) - \int_{\{0\}} f(x) = f(1) - f(0). $$
I hope this application convinces you that there should be a minus sign, and that the above theory produced something sensible.
