Here's an answer to the modified question in the comments:
Question: Let $X$ be an affine scheme, and $\mathcal F$ an $\mathcal O_X$-module. Under what conditions is $\mathcal F$ quasicoherent?
As observed by Thomason-Trobaugh (see Appendix B of "Higher Algebraic K-Theory of Schemes" in the Grothendieck Festschrift), for reasonable schemes the quasicoherent sheaves are coreflective in the category of $\mathcal O_X$-modules. For affine schemes, the coreflection is particularly easy to describe: it sends $\mathcal F$ to the sheafification of $\bar {\mathcal F}: U \mapsto \mathcal F(X) \otimes_{\mathcal O_X(X)} \mathcal O_X(U)$. That is,
Answer: $\mathcal F$ is quasicoherent if and only if the canonical map $\bar{\mathcal F} \to \mathcal F$ is an isomorphism after sheafification.
This can be checked on stalks without taking an explicit sheafification. For $p \in X$, the map $\bar {\mathcal F}_p \to \mathcal F_p$ is the map
$$\varinjlim_{U \ni p} \mathcal F(X) \otimes_{\mathcal O_X(X)} \mathcal O_X(U) \to \varinjlim_{U \ni p} \mathcal F(U)$$
- This map is surjective if and only if for all $U \ni p$ and for all $f \in \mathcal F(U)$ there exist $g_1,\dots, g_n \in \mathcal F(X)$, $U \supseteq V \ni p$, and $\varphi_1,\dots, \varphi_n \in \mathcal O_X(V)$ such that $f|_V = \sum_{i=1}^n \varphi_i g_i|_V$.
The map is injective if and only if for all $g_1,\dots, g_n \in \mathcal F(X)$, $U \ni p$ and $\varphi_1,\dots, \varphi_n \in \mathcal O_X(U)$, if $\sum_{i=1}^n \varphi_i g_i = 0$ in $\mathcal F(U)$, then for some $U \supseteq V \ni p$, we have $\sum_{i=1}^n \varphi_i|_V g_i = 0$ in $\mathcal F(X) \otimes_{\mathcal O_X(X)} \mathcal O_X(V)$.
At least if $X$ is irreducible, these conditions may be rephrased as follows:
At least if $X$ is irreducible, another way of phrasing the surjectivitySurjectivity: condition is the following. For all $U \ni p$ and all $f \in \mathcal F(U)$, there exists $g \in \mathcal F(X)$ and $\psi \in \mathcal O_X(X)$ not vanishing at $p$ such that $\psi|_V f|_V = g|_V$ for some $U \supseteq V \ni p$$\psi|_U f = g|_U$.
At least if $X$ is irreducible, another way of phrasing the injectivityInjectivity: condition is the following: forFor all $f \in \mathcal F(X)$ and $U \ni p$, if $f|_U = 0$ in $\mathcal F(U)$, then there exists $\psi \in \mathcal O_X(X)$ not vanishing at $p$ such that $\psi f= 0$ in $\mathcal F(X)$.