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I think, the answer can depend on how you interpret the question. Let me show that the answer is negative for one of the interpretations already in the case of $2$-dimensional manifolds. We study the question locally in a neighborhood neighbourhood of a vertex of a triangulation, so the conidtion condition on the integral over $n$-simplexes does not play any role. The obstruction for the exitense existence of $\beta$ comes form the locall local behaviour of curves at a vertex.

Lemma. Let $\alpha=dx\wedge dy$ on $\mathbb R^2$. For $n\ge 8$ there exist $\gamma_1,...,\gamma_n$, smooth rays on $\mathbb R^2$ that meet at $0$ with different tangent vectors and such that there is no $\beta$ defined in any neighbourhood of 0 with $d\beta=\alpha$ and vanishing been resticted restricted to $\gamma_i$.

It is clear that this lemma implies the negative answer to a version of the question, when we are not allowd allowed to deform the triangulation.

Proof of Lemma. Suppose by contradcition contradiction that $\beta$ exists. Then $\beta_1=\beta-\frac{1}{2}(xdy-ydx)$ is a closed 1-form. So we can write $\beta_1=dF$, where $F$ is a function defined in a neighborhood neighbourhood of $0$, $F(0)=0$. Since the number $n$ of the rays is more than $2$, $dF$ should vanish at zero. Moreover, it is not hard to see, that since the number of rays is more than $4$, the quadratic term of $F$ vanishes at zero too.

Now, since $\beta$ vanishes on $\gamma_i$, the restiction of $\beta_1$ to $\gamma_i$ equals $\frac{1}{2}(ydx-xdy)$. So we get the formula for $F$, resticted to $\gamma_i$ $$F=\frac{1}{2}\int_{\gamma_i}ydx-xdy.$$ Now, we will chose the rays $\gamma_1,...,\gamma_8$. Namely $\gamma_1(t)=(t,t^2)$, $\gamma_2(t)=(t,t-t^2)$, and take $\gamma_3,...,\gamma_8$ by consequtively consecutively rotating $\gamma_1,\gamma_2$ by $\pi/2$, $\pi$, $3\pi/2$.

It is not hard to see, that $F$ is cubic modulo higher terms in $t$ when it is resticted restricted to $\gamma_i$. At the same time $F$ is positive on $\gamma_{1},\gamma_3,\gamma_5, \gamma_7$ and negative resticted restricted to other rays. So it changes its sign at lest $8$ times on a little circle surrounding $0$. This is impossible for a cubic Function (in a little neighborhood neighbourhood the cubic term of $F$ should be dominating). Contradiction.

2 added 237 characters in body

I think, the answer can depend on how you interpret the question. Let me show that the answer is negative for one of the interpretations already in the case of $2$-dimensional manifolds. We study the question locally in a neighborhood of a vertex of a triangulation, so the conidtion on the integral does not play any role. The obstruction for the exitense of $\beta$ comes form the locall behaviour of curves at a vertex.

Lemma. Let $\alpha=dx\wedge dy$ on $\mathbb R^2$. For $n\ge 8$ there exist $\gamma_1,...,\gamma_n$, smooth rays on $\mathbb R^2$ that meet at $0$ with different tangent vectors and such that there is no $\beta$ with $d\beta=\alpha$ and vanishing been resticted to $\gamma_i$.

It is clear that this lemma implies the negative answer to a version of the question, when we are not allowd to deform the triangulation.

Proof of Lemma. Suppose by contradcition that $\beta$ exists. Then $\beta_1=\beta-\frac{1}{2}(xdy-ydx)$ is a closed 1-form. So we can write $\beta_1=dF$, where $F$ is a function defined in a neighborhood of $0$, $F(0)=0$. Since the number $n$ of the rays is more than $2$, $dF$ should vanish at zero. Moreover, it is not hard to see, that since the number of rays is more than $4$, the quadratic term of $F$ vanishes at zero too.

Now, since $\beta$ vanishes on $\gamma_i$, the restiction of $\beta_1$ to $\gamma_i$ equals $\frac{1}{2}(ydx-xdy)$. So we get the formula for $F$, resticted to $\gamma_i$ $$F=\frac{1}{2}\int_{\gamma_i}ydx-xdy.$$

Now, we will chose the rays $\gamma_1,...,\gamma_8$. Namely $\gamma_1(t)=(t,t^2)$, $\gamma_2(t)=(t,t-t^2)$, and take $\gamma_3,...,\gamma_8$ by consequtively rotating $\gamma_1,\gamma_2$ by $\pi/2$, $\pi$, $3\pi/2$.

It is not hard to see, that $F$ is cubic modulo higher terms in $t$ when it is resticted to $\gamma_i$. At the same time $F$ is positive on $\gamma_{1},\gamma_3,\gamma_5, \gamma_7$ and negative resticted to other rays. So it changes its sign at lest $8$ times on a little circle surrounding $0$. This is impossible for a cubic Function (in a little neighborhood the cubic term of $F$ should be dominating). Contradiction.

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I think, the answer can depend on how you interpret the question. Let me show that the answer is negative for one of the interpretations already in the case of $2$-dimensional manifolds.

Lemma. Let $\alpha=dx\wedge dy$ on $\mathbb R^2$. For $n\ge 8$ there exist $\gamma_1,...,\gamma_n$, smooth rays on $\mathbb R^2$ that meet at $0$ with different tangent vectors and such that there is no $\beta$ with $d\beta=\alpha$ and vanishing been resticted to $\gamma_i$.

It is clear that this lemma implies the negative answer to a version of the question, when we are not allowd to deform the triangulation.

Proof of Lemma. Suppose by contradcition that $\beta$ exists. Then $\beta_1=\beta-\frac{1}{2}(xdy-ydx)$ is a closed 1-form. So we can write $\beta_1=dF$, where $F$ is a function defined in a neighborhood of $0$, $F(0)=0$. Since the number $n$ of the rays is more than $2$, $dF$ should vanish at zero. Moreover, it is not hard to see, that since the number of rays is more than $4$, the quadratic term of $F$ vanishes at zero too.

Now, since $\beta$ vanishes on $\gamma_i$, the restiction of $\beta_1$ to $\gamma_i$ equals $\frac{1}{2}(ydx-xdy)$. So we get the formula for $F$, resticted to $\gamma_i$ $$F=\frac{1}{2}\int_{\gamma_i}ydx-xdy.$$

Now, we will chose the rays $\gamma_1,...,\gamma_8$. Namely $\gamma_1(t)=(t,t^2)$, $\gamma_2(t)=(t,t-t^2)$, and take $\gamma_3,...,\gamma_8$ by consequtively rotating $\gamma_1,\gamma_2$ by $\pi/2$, $\pi$, $3\pi/2$.

It is not hard to see, that $F$ is cubic modulo higher terms in $t$ when it is resticted to $\gamma_i$. At the same time $F$ is positive on $\gamma_{1},\gamma_3,\gamma_5, \gamma_7$ and negative resticted to other rays. So it changes its sign at lest $8$ times on a little circle surrounding $0$. This is impossible for a cubic Function (in a little neighborhood the cubic term of $F$ should be dominating). Contradiction.