This answer is based on the comment above and Lemma 2.9 of the thesis of Ajay, as mentioned in that comment. Up to minor adaptation, the result proved in the Lemma can be stated as follows
Lemma: For $z \in \mathbb{R}^n$, let $T_z: \mathcal{D}(\mathbb{R}^m \times \mathbb{R}^n) \to \mathcal{D}(\mathbb{R}^m \times \mathbb{R}^n)$ be defined by $T_zf(x,y) = f(x, y+z)$ where $x \in \mathbb{R}^m, y \in \mathbb{R}^n$. Suppose that $\phi \in \mathcal{D}'(\mathbb{R}^m \times \mathbb{R}^n)$ has the property that $T_z^*\phi = \phi$ for each $z \in \mathbb{R}^n$. Then there is a $\psi \in \mathcal{D}'(\mathbb{R}^m)$ such that
$$\phi(f) = \psi \bigg( \int_{\mathbb{R}^n} f(\cdot, y) dy \bigg).$$
To prove this, first note that it is sufficient to prove the case $n = 1$ since iterating this argument will give the result for general $n$. The key is then to note that $\phi$ vanishes on test functions which are of the form $\partial_{x_{m+1}}g$ for some $g$, by translation invariance. This means that if we pick $h \in \mathcal{D}(\mathbb{R})$ with $\int h = 1$ and set $\psi(g) = \phi(g \otimes h)$ then we have
$$\phi(f) - \psi\bigg( \int_{\mathbb{R}} f(\cdot, y) dy \bigg) = \phi(f - \tilde{f})$$
where $\tilde{f}(x,y) = h(y) \int f(x,z) dz$. Then since $\int f(\cdot, z) - \tilde{f}(\cdot,z) dz = 0$, it is a standard fact that there is a $g$ such that $\partial_{x_{m+1}}g = f - \tilde{f}$ which implies that $\phi(f-\tilde{f}) = 0$ so that
$$\phi(f) = \psi\bigg(\int f(\cdot,y) dy \bigg)$$
as desired.
To conclude the result of the question, we just need to make a change of variables. Define $\tau(x,y) = (x-y,-y)$ and let $\tilde{F} = F \circ \tau^*$. Then $\tilde{F}(T_zf) = F((T_zf) \circ \tau)$. If $\tau_z f(x,y) = f(x-z,y-z)$ then we have that $(T_zf) \circ \tau = \tau_z (f \circ \tau)$ so by translation invariance of $F$, $\tilde{F}(T_zf) = \tilde{F}(f)$. Hence, by the Lemma there is an $\mathcal{R}[F]$ such that
$$\tilde{F}(f) = \mathcal{R}[F]\bigg( \int_{\mathbb{R}^n} f(\cdot, y) dy \bigg).$$
Then, given $f$, $F(f) = \tilde{F}(\tau^*f)$ since $\tau^2 = \operatorname{Id}$. Hence
$$F(f) = \mathcal{R}[F]\bigg(\int_{\mathbb{R}^n} \tau^*f(\cdot, y) dy \bigg)$$
which is exactly the desired result when $f$ is a tensor product.
Best Answer
You're misunderstanding translation-invariance here.
A translation-invariant operator $T \colon L^p(\mathbb{R}^d) \to L^q(\mathbb{R}^d)$ is an operator that commutes with translations, i.e. $T \circ \tau_c = \tau_c \circ T$ for all $c \in \mathbb{R}^d$.
Then, with a bit of ho-humming (that can be made precise, see below), when the $x_k$ are spread far enough apart, the $T(\tau_{x_k}g) = \tau_{x_k}Tg$ have the bulk of their weights separated, so
$$\lVert Tf\rVert_q = \lVert \sum \tau_{x_k}Tg\rVert_q \approx \left(\int \sum \left\lvert (Tg)(x-x_k)\right\rvert^q\,dx \right)^{1/q} \approx \left(\sum \int \lvert(Tg)(x-x_k)\rvert^q\,dx \right)^{1/q} \approx N^{1/q} \lVert Tg\rVert_q$$
For $f$ itself, with the assumption of compact support there is no problem seeing $\lVert f\rVert_p = N^{1/p}\lVert g\rVert_p$.
To conclude that that implies $q \geqslant p$, one needs $T \neq 0$.
Making the ho-humming precise:
Let
$$\chi_r(x) = \begin{cases} 1,\quad \lVert x\rVert \leqslant r\\ 0,\quad \lVert x\rVert > r. \end{cases}$$
Fix a (smooth) $g \in L^p(\mathbb{R}^d)$ with compact support in $B_R(0)$.
Then, for $f = \sum\limits_{k = 1}^N \tau_{x_k}g$ and $r > 0$, if $\lVert x_i - x_j\rVert \geqslant 2r$ for all $i \neq j$, you have
$$\begin{align} \lVert Tf\rVert_q &= \left\lVert\left(\sum_{k=1}^N \tau_{x_k}\bigl(\chi_r\cdot(Tg)\bigr)\right) + \left(\sum_{k=1}^N \tau_{x_k}\bigl((1-\chi_r)\cdot(Tg)\bigr)\right) \right\rVert_q\\ &\leqslant \left\lVert\left(\sum_{k=1}^N \tau_{x_k}\bigl(\chi_r\cdot(Tg)\bigr)\right)\right\rVert_q + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q\\ &= \left(\int \sum_{k=1}^N \left\lvert\chi_r(x-x_k)(Tg)(x-x_k)\right\rvert^q\,dx\right)^{1/q} + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q\\ &= \left(N\cdot \lVert \chi_r\cdot (Tg)\rVert^q\right)^{1/q} + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q\\ &= N^{1/q} \lVert \chi_r\cdot (Tg)\rVert_q + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q\\ &\leqslant N^{1/q} \lVert Tg\rVert_q + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q. \end{align}$$
The first inequality follows from the triangle inequality for $\lVert\cdot\rVert_q$. The next equalities follow from the disjointness of the supports of the $\tau_{x_k}\bigl(\chi_r\cdot(Tg)\bigr)$ and the translation-invariance (note: since the result is a number, and not a function, translation-invariance means $\lambda(\tau_c M) = \lambda(M)$ for all $c$ and measurable $M$ here) of the Lebesgue measure. The final inequality from $\lVert\chi_r\cdot h\rVert_q \leqslant \lVert h\rVert_q$ for all $h$.
Using the triangle inequality in the form $\lVert a + b \rVert \geqslant \lVert a\rVert - \lVert b\rVert$ for the split between the $\chi_r(Tg)$ and $(1-\chi_r)(Tg)$, we obtain
$$\lVert Tf\rVert_q \geqslant N^{1/q} \lVert \chi_r\cdot (Tg)\rVert_q - N\cdot \lVert (1-\chi_r)(Tg)\rVert_q$$
Now, given $g$, $N$, and an arbitrary $\varepsilon > 0$, by the dominated convergence theorem, you can choose $r_0 > 0$ so that $\lVert (1-\chi_r)\cdot(Tg)\rVert_q < \varepsilon/N$ for all $r \geqslant r_0$. Choose additionally $r_0 > 2R$, so that the $\tau_{x_k}g$ have disjoint support.
If the $x_k$ are then chosen far enough apart, you find
$$N^{1/q}\lVert Tg\rVert_q - \frac{\varepsilon}{N^{1-1/q}} - N\cdot\frac{\varepsilon}{N} \leqslant \lVert Tf\rVert_q \leqslant N^{1/q}\lVert Tg\rVert_q + \varepsilon,$$
so $\lVert Tf\rVert_q \approx N^{1/q}\lVert g\rVert_q$, and
$$N^{1/q}\lVert Tg\rVert_q - 2\varepsilon \leqslant \lVert Tf\rVert_q \leqslant \lVert T\rVert \cdot \lVert f\rVert_p = N^{1/p}\lVert T\rVert\cdot\lVert g\rVert_p.$$
Since $\varepsilon$ could be arbitrarily chosen,
$$N^{1/q}\lVert Tg\rVert_q \leqslant N^{1/p}\lVert T\rVert\cdot \lVert g\rVert_p$$
for all $N$ and (smooth) $g$ with compact support. If $T \neq 0$, a smooth $g$ with compact support and $Tg \neq 0$ exists ($\mathscr{C}_c(\mathbb{R}^d)$ is dense in $L^p(\mathbb{R}^d)$ for $p < \infty$). Then, taking the limit for $N \to \infty$ would lead to a contradiction if $p > q$.