[proofplan]
We replace a neighbourhood of $N$ in $M$ by a fibrewise star-shaped tubular neighbourhood in the normal bundle of $N$. On this tubular neighbourhood, the radial contraction onto the zero section gives a homotopy operator $K$ for differential forms. The homotopy formula expresses a form as the sum of its pullback from the zero section and an exact term, up to the [exterior derivative](/theorems/1525) of the form. Since $\alpha$ is closed and restricts to zero on $N$, this formula produces the desired primitive, and the radial vector field vanishes on the zero section, giving the relative condition on $\beta$.
[/proofplan]
[step:Choose a fibrewise star-shaped tubular neighbourhood inside $V$]
Choose a [smooth vector bundle](/page/Smooth%20Vector%20Bundle) $\nu\to N$ complementary to $TN$ in $TM|_N$, and let $\pi_\nu:\nu\to N$ denote its bundle projection. By the [Tubular Neighbourhood Theorem](/theorems/2276) for closed embedded submanifolds, there are an open neighbourhood $E_0\subset\nu$ of the zero section and a smooth embedding
\begin{align*}
\Phi:E_0\to M
\end{align*}
whose image is an open neighbourhood of $N$ in $M$ and whose restriction to the zero section is the inclusion $i:N\to M$.
Since $V\subset M$ is open and contains $N$, we shrink $E_0$ to a fibrewise star-shaped open neighbourhood $E\subset E_0$ of the zero section such that $\Phi(E)\subset V$. Concretely, after choosing a smooth bundle metric on $\nu$, one may take
\begin{align*}
E=\{v\in\nu: |v|_{\nu}<\rho(\pi_\nu(v))\}
\end{align*}
for a positive smooth function $\rho:N\to(0,\infty)$ chosen small enough that this set lies in $\Phi^{-1}(V)$. This choice is possible because $N$ is paracompact as a closed embedded submanifold of the second-countable Hausdorff manifold $M$.
Define
\begin{align*}
U:=\Phi(E)\subset V.
\end{align*}
Then $U$ is an open neighbourhood of $N$ in $V$, and $\Phi:E\to U$ is a diffeomorphism. Let
\begin{align*}
z:N\to E
\end{align*}
be the zero-section inclusion. Then $\Phi\circ z=i_U$ and $\pi_\nu\circ z=\operatorname{id}_N$.
[/step]
[step:Construct the radial homotopy operator on the tubular neighbourhood]
Define the radial contraction
\begin{align*}
H:[0,1]\times E\to E,\qquad H(t,v)=t v.
\end{align*}
The fibrewise star-shaped choice of $E$ ensures that $t v\in E$ for every $v\in E$ and every $t\in[0,1]$. Write $H_t:E\to E$ for the smooth map $H_t(v)=t v$. Thus $H_1=\operatorname{id}_E$ and $H_0=z\circ\pi_\nu$.
For $\gamma\in\Omega^r(E)$ with $r\ge 1$, define the homotopy operator
\begin{align*}
K\gamma\in\Omega^{r-1}(E)
\end{align*}
by
\begin{align*}
(K\gamma)_v(w_1,\dots,w_{r-1})=\int_0^1 \gamma_{H_t(v)}\bigl(\partial_tH(t,v),d(H_t)_v(w_1),\dots,d(H_t)_v(w_{r-1})\bigr)\,d\mathcal L^1(t).
\end{align*}
Here $v\in E$, the vectors $w_1,\dots,w_{r-1}$ lie in $T_vE$, and $\partial_tH(t,v)\in T_{H_t(v)}E$ denotes the velocity of the curve $t\mapsto H_t(v)$. For $r=0$, set $K\gamma=0$.
The homotopy formula for differential forms gives, for every $\gamma\in\Omega^r(E)$,
\begin{align*}
H_1^*\gamma-H_0^*\gamma=dK\gamma+K(d\gamma).
\end{align*}
Since $H_1=\operatorname{id}_E$ and $H_0=z\circ\pi_\nu$, this becomes
\begin{align*}
\gamma-\pi_\nu^*z^*\gamma=dK\gamma+K(d\gamma).
\end{align*}
[guided]
The point of passing to the tubular neighbourhood is that there is now a canonical deformation back to $N$: each normal vector $v\in\nu$ can be scaled to $t v$. The fibrewise star-shaped condition on $E$ is exactly what makes the map
\begin{align*}
H:[0,1]\times E\to E,\qquad H(t,v)=t v
\end{align*}
well-defined on the whole interval $[0,1]$.
For each $t\in[0,1]$, define
\begin{align*}
H_t:E\to E,\qquad H_t(v)=t v.
\end{align*}
Then $H_1=\operatorname{id}_E$, while $H_0$ sends every vector $v\in E$ to the zero vector in the fibre over $\pi_\nu(v)$. Therefore
\begin{align*}
H_0=z\circ\pi_\nu.
\end{align*}
The homotopy operator lowers degree by one. If $\gamma\in\Omega^r(E)$ with $r\ge 1$, define $K\gamma\in\Omega^{r-1}(E)$ by
\begin{align*}
(K\gamma)_v(w_1,\dots,w_{r-1})=\int_0^1 \gamma_{H_t(v)}\bigl(\partial_tH(t,v),d(H_t)_v(w_1),\dots,d(H_t)_v(w_{r-1})\bigr)\,d\mathcal L^1(t).
\end{align*}
The vector $\partial_tH(t,v)$ is inserted into the first slot of $\gamma$, and the remaining vectors are pushed forward by $d(H_t)_v$. The integral is taken with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal L^1$ on $[0,1]$. For $r=0$, we set $K\gamma=0$ because there is no form of degree $-1$.
The homotopy formula for differential forms states that this operator satisfies
\begin{align*}
H_1^*\gamma-H_0^*\gamma=dK\gamma+K(d\gamma).
\end{align*}
Substituting $H_1=\operatorname{id}_E$ and $H_0=z\circ\pi_\nu$ gives
\begin{align*}
\gamma-\pi_\nu^*z^*\gamma=dK\gamma+K(d\gamma).
\end{align*}
This identity is the algebraic heart of the relative Poincare lemma: it says that a form differs from its restriction to the zero section by an exact term, with an additional error term involving its exterior derivative.
[/guided]
[/step]
[step:Apply the homotopy formula to the pulled back closed form]
Define
\begin{align*}
\widetilde{\alpha}:=\Phi^*(\alpha|_U)\in\Omega^k(E).
\end{align*}
Since exterior differentiation commutes with pullback and $d\alpha=0$ on $V$,
\begin{align*}
d\widetilde{\alpha}=d\Phi^*(\alpha|_U)=\Phi^*(d(\alpha|_U))=0.
\end{align*}
Also, because $\Phi\circ z=i_U$ and $i_U^*(\alpha|_U)=i_V^*\alpha=0$,
\begin{align*}
z^*\widetilde{\alpha}=z^*\Phi^*(\alpha|_U)=(\Phi\circ z)^*(\alpha|_U)=i_U^*(\alpha|_U)=0.
\end{align*}
Applying the homotopy formula to $\gamma=\widetilde{\alpha}$ gives
\begin{align*}
\widetilde{\alpha}-\pi_\nu^*z^*\widetilde{\alpha}=dK\widetilde{\alpha}+K(d\widetilde{\alpha}).
\end{align*}
The two terms involving $z^*\widetilde{\alpha}$ and $d\widetilde{\alpha}$ vanish, so
\begin{align*}
\widetilde{\alpha}=dK\widetilde{\alpha}.
\end{align*}
[/step]
[step:Push the primitive back to $U$ and verify it differentiates to $\alpha$]
Define
\begin{align*}
\beta:=(\Phi^{-1})^*(K\widetilde{\alpha})\in\Omega^{k-1}(U).
\end{align*}
Since exterior differentiation commutes with pullback by the diffeomorphism $\Phi^{-1}:U\to E$,
\begin{align*}
d\beta=d(\Phi^{-1})^*(K\widetilde{\alpha})=(\Phi^{-1})^*(dK\widetilde{\alpha})=(\Phi^{-1})^*\widetilde{\alpha}.
\end{align*}
By the definition of $\widetilde{\alpha}$,
\begin{align*}
(\Phi^{-1})^*\widetilde{\alpha}=(\Phi^{-1})^*\Phi^*(\alpha|_U)=\alpha|_U.
\end{align*}
Hence
\begin{align*}
d\beta=\alpha|_U.
\end{align*}
[/step]
[step:Check that the primitive vanishes along the submanifold]
It remains to verify the relative condition. Since $i_U=\Phi\circ z$,
\begin{align*}
i_U^*\beta=z^*\Phi^*(\Phi^{-1})^*(K\widetilde{\alpha})=z^*(K\widetilde{\alpha}).
\end{align*}
For every $x\in N$, the path $t\mapsto H_t(z(x))$ is constant equal to $z(x)$, so
\begin{align*}
\partial_tH(t,z(x))=0\in T_{z(x)}E.
\end{align*}
The defining formula for $K$ therefore gives
\begin{align*}
(z^*K\widetilde{\alpha})_x=0
\end{align*}
as a $(k-1)$-form on $T_xN$. When $k=1$, this says the resulting smooth function has value $0$ at every point of $N$. Thus
\begin{align*}
i_U^*\beta=0.
\end{align*}
The form $\beta\in\Omega^{k-1}(U)$ therefore satisfies both required conclusions.
[/step]
