OK, here is the full story, which confirms what is written in Daniel Bruegmann's answer. In what follows, I will work with a fixed prefactorization algebra $F$ and an $n$-manifold $M$. I will make use of the fact that the homotopy colimits of simplicial (co)chain complexes are given by the (direct sum) totalization of the associated double complex; see my note [Ara] for a detailed account on this. Our argument will also make use of the fact that reindexing by homotopy final functors preserves homotopy colimits; [ArØr] is a good reference for this.
A FEW REMARKS
Definition (2) does not quite make sense. Indeed, as soon as $\mathcal{U}$ contains some element $U$, we would need a map $I=F(\emptyset,\{U\})\to F(\{U\})$ corresponding to the face map $d_{0}$; but I don't know where such a map comes from (except when $F(\emptyset)=I$). I will therefore redefine $F(\alpha_{0},\dots,\alpha_{n})$ in Definition (2) as $$ F(\alpha_{0},\dots,\alpha_{n})=F\left(\bigcap_{i=0}^{n}O(\alpha_{i})\right), $$ where we set $O(\alpha)=\bigcup_{V\in\alpha}V$ for each $\alpha\in P\mathcal{U}$.
In Definition (3), we should index the homotopy colimits by {}nonempty{} subsets of $\mathcal{U}$. Otherwise, the homotopy colimit will just be $FU$.
THE EQUIVALENCES
We will prove the following:
Proposition 1. Let $F$ be a prefactorization algebra on $M$. Suppose that, for each finite set $U_{1},\dots,U_{k}$ of open sets of $M$, the map $\bigotimes_{i=1}^{k}F(U_{i})\to F(\bigcup_{i=1}^{k}U_{i})$ is a quasi-isomorphism. Then the following conditions are equivalent:
(A) $F$ is a factorization algebra in the sense of Definition (1).
(B) $F$ is a factorization algebra in the sense of Definition (2) (or rather, its modified version explained in the above remark)
(C) $F$ is a factorization algebra in the sense of Definition (3).
(D) Let $\mathcal{U}$ be a Weiss cover of an open set $U\subset M$ such that, for each $V\in\mathcal{U}$, every open subset of $V$ belongs to $\mathcal{U}$. Then the map $$ \operatorname{hocolim}_{V\in\mathcal{U}}F(V)\to F(U) $$ is a quasi-isomorphism.
We need a lemma:
Lemma 2. Let $S$ be a nonempty set. Let $\mathcal{X}(S)$ denote the category whose objects are the finite nonempty sequences $\left(x_{0},\dots,x_{n}\right)$ of elements of $S$, and whose morphism $\left(x_{0},\dots,x_{n}\right)\to\left(y_{0},\dots,y_{m}\right)$ is a poset map $u:[m]\to[n]$ such that $x_{u(i)}=y_{i}$ for every $0\leq i\leq m$. The nerve of $\mathcal{X}(S)$ is weakly contractible.
(Proof of Lemma 2.) The category $\mathcal{X}(S)$ is the category of elements of the functor $S^{[\bullet]}:\mathbf{\Delta}^{\mathrm{op}}\to\mathsf{Set}$ defined by $[n]\mapsto S^{[n]}$. By Thomason's homotopy colimit theorem, the simplicial set $N(\mathcal{C})$ has the weak homotopy type of the homotopy colimit of the composite $\mathbf{\Delta}^{\mathrm{op}}\xrightarrow{S^{[\bullet]}}\mathsf{Set}\hookrightarrow\mathsf{sSet}$. The latter is given by the diagonal of the corresponding bisimplicial set, which, in this case, is the simplicial set $S^{[\bullet]}$. It will therefore suffice to show that the simplicial set $S^{[\bullet]}$ is weakly contractible. We claim that $S^{[\bullet]}$ is a contractible Kan complex. Observe that, given a simplicial set $X$, defining a map of simplicial sets $X\to S^{[\bullet]}$ is equivalent to defining a set map $X_{0}\to S$. So the simplicial set has the extension property for the inclusion $\partial\Delta^{n}\subset\Delta^{n}$ for every $n>0$. Since $S$ is nonempty, the simplicial set $S^{[\bullet]}$ has the extension property for the inclusion $\partial\Delta^{0}\subset\Delta^{0}$. $\blacksquare$
(Proof of Proposition 1.)
(A)$\implies$(D): Let $\mathcal{U}$ be as in (D). The simplicial object $\left\{ \bigoplus_{U_{0},\dots,U_{n}\in\mathcal{U}}F\left(\bigcap_{i=0}^{n}U_{i}\right)\right\} _{n\geq0}$ is a left Kan extension of the composite $\mathcal{X}(\mathcal{U})\xrightarrow{\varphi}\mathcal{U}\xrightarrow{F}\mathcal{C}$ along the projection $\mathcal{X}\to\mathbf{\Delta}^{\mathrm{op}}$, where the functor $\varphi$ is given by $\varphi(U_{0},\dots,U_{n})=\bigcap_{i=0}^{n}U_{i}$. Therefore, condition (A) implies that the map $\operatorname{hocolim}_{(U_{0},\dots,U_{n})\in\mathcal{X}}F(\bigcap_{i=0}^{n}U_{i})\to FU$ is a quasi-isomorphism. It will therefore suffice to show that $\varphi$ is homotopy final, which follows from Lemma 2.
(D)$\implies$(A): Let $\mathcal{U}$ be a Weiss cover of an open set $U\subset M$. Arguing as in the previous paragraph, it will suffice to show that the map $\operatorname{hocolim}_{(U_{0},\dots,U_{n})\in\mathcal{X}}F(\bigcap_{i=0}^{n}U_{i})\to FU$ is a quasi-isomorphism. Let $\overline{\mathcal{U}}$ denote the set of open subsets of elements of $\mathcal{U}$, and define $\varphi:\mathcal{X}(\mathcal{U})\to\overline{\mathcal{U}}$ as in the previous paragraph. By Lemma 2, the functor $\varphi$ is homotopy final, so condition (D) implies that the map $\operatorname{hocolim}_{(U_{0},\dots,U_{n})\in\mathcal{X}}F(\bigcap_{i=0}^{n}U_{i})\to FU$ is a quasi-isomorphism, as claimed.
(D)$\implies$(C): Let $\mathcal{U}$ be a Weiss cover of an open set $U\subset M$. Let $P$ denote the set of finite nonempty subsets of $\mathcal{U}$, and let $\overline{\mathcal{U}}$ denote the set of open subsets of elements of $\mathcal{U}$. The assignment $\alpha\mapsto\bigcap_{V\in\alpha}V$ determines a map $\varphi:P^{\mathrm{op}}\to\overline{\mathcal{U}}$. To complete the proof, it suffices to show that $\varphi$ is homotopy final, which is trivial.
(C)$\implies$(D): This is similar to (C)$\implies$(D).
(B)$\implies$(D): Let $\mathcal{U}$ be as in (D). Arguing as in the proof of (A)$\implies$(D) (and noting that $\mathcal{U}$ is a factorizing cover), we are reduced to showing that the functor $\mathcal{X}(P\mathcal{U})\to\mathcal{U}$ defined by $(\alpha_{0},\dots,\alpha_{n})\mapsto\bigcap_{i=0}^{n}O(\alpha_{i})$ is homotopy final. This follows from Lemma 2.
(D)$\implies$(B): Let $\mathcal{U}$ be a factorizing cover of an open set $U\subset M$. Let $\mathcal{V}$ denote the set of all open subsets of the topological spaces $\{U(\alpha)\}_{\alpha\in P\mathcal{U}}$. Note that $\mathcal{V}$ is a Weiss cover. Arguing as in the proof of (D)$\implies$(A), we are reduced to showing that the functor $\mathcal{X}(P\mathcal{U})\to\mathcal{V}$ defined by $(\alpha_{0},\dots,\alpha_{n})\mapsto\bigcap_{i=0}^{n}O(\alpha_{i})$ is homotopy final. This is a consequence of Lemma 2. $\blacksquare$
REFERENCES
[Ara] Arakawa, K. (2023). Homotopy Limits and Homotopy Colimits of Chain Complexes. arxiv.2310.00201
[ArØr] Arkhipov, S., Ørsted, S. Homotopy (co)limits via homotopy (co)ends in general combinatorial model categories. arxiv.1807.03266