Let $n\in\mathbb{N}$, $f \in F_1(\mathbb{R}^n)$, say, $$f = \sum_{j=1,...,n}f_jdx_j,$$ with $df=0$. Define for each $x\in\mathbb{R}^n$, $$u(x)=\sum_{j=1,...,n}\int_{[0,1]}f_j(tx)p_j(x)dt$$ ($p_j$'s are projections). Then $u \in F_0(\mathbb{R}^n)$ and $du=f$.
It seems to me that it is clear $u$ is a $0$-form, correct? Now, what I (think I) can do so far is $$du = d\sum_{j=1,...,n}\int_{[0,1]}f_j(t-)p_j(-)dt = \sum_{j=1,...,n}d\int_{[0,1]}f_j(t-)p_j(-)dt =$$ $$ \sum_{j=1,...,n}\sum_{k=1,...,n}\partial_k(\int_{[0,1]}f_j(t-)p_j(-)dt)dx_k = \sum_{j=1,...,n}\sum_{k=1,...,n}\int_{[0,1]}\partial_k(f_j(t-)p_j(-))dtdx_k =$$ $$\sum_{j=1,...,n}\sum_{k=1,...,n}\int_{[0,1]}(t(\partial_k(f_j))(t-)p_j(-)+f_j(t-)\partial_k(p_j)(-))dtdx_k = ...$$
am I still good here, or did I go astray? If I'm good, how do I continue? Am I making things more complicated than they are?
Also, I realize that the hypothesis is that $$0 = df = \sum_{j=1,...,n}\sum_{k=1,...,n} \partial_k(f_j) dx_k \wedge dx_j.$$
Thanks in advance.
Firstly, your $f$ is a $1$-form and not a $0$-form, so I'd guess you meant $f \in F_1(\Bbb R^n)$ (a more common notation for $F_k$ is $\Omega^k$, anyway). Yes, your $u$ is a $0$-form, as $0$-forms are just smooth functions (taking zero inputs).
Note that ${\rm d}f = 0$ is equivalent to the condition $\partial f_j/\partial x_i = \partial f_i/\partial x_j$ for every possibilities of $i$ and $j$. I will use this in the step in red below. We want to check that $\partial u/\partial x_i = f_i$.
Just compute $$\begin{align}\frac{\partial u}{\partial x_i}(x) &= \frac{\partial}{\partial x_i} \sum_{j=1}^n \int_0^1x_jf_j(tx)\,dt \\ &= \sum_{j=1}^n \int_0^1 \frac{\partial x_j}{\partial x_i}f_j(tx)+ x_j \frac{\partial}{\partial x_i} (f_j(tx))\,dt \\ &= \sum_{j=1}^n \int_0^1 \delta_{ij}f_j(tx) + x_j\left(t\frac{\partial f_j}{\partial x_i}(tx)\right)\,dt\\ &= \int_0^1 f_i(tx)\,dt+ \int_0^1 t \left(\sum_{j=1}^n x_j \frac{\partial f_j}{\partial x_i}(tx)\right)\,dt \\ &= \int_0^1 f_i(tx)\,dt+ \int_0^1 t \left(\sum_{j=1}^n x_j \frac{\partial f_{\color{red}{i}}}{\partial x_{\color{red}{j}}}(tx)\right)\,dt .\end{align}$$Now, we integrate what's left by parts, noting that $tf_i(tx)\big|_0^1 = f_i(x)$: $$\frac{\partial u}{\partial x_i}(x) = \int_0^1f_i(tx) \,dt + f_i(x) - \int_0^1f_i(tx) \,dt = f_i(x),$$as wanted. The integration by parts is done bearing in mind that $$\sum_{j=1}^n x_j \frac{\partial f_i}{\partial x_j}(tx) = \frac{d}{dt} (f_i(tx)).$$