As pointed out in the comments, the correct definition of “the sections $s_1,\ldots,s_n$ (on $X$, say) generate $\mathcal{F}$ [as an $\mathcal{A}$-module]” is that the morphism of $\mathcal{A}$-modules $\mathcal{A}^{\oplus n} \to \mathcal{F}$ they define is surjective, so the question that remains is whether this is equivalent to the definition originally proposed in the question, which all boils down to:
If global sections $s_1,\ldots,s_n$ generate $\mathcal{F}$ as an $\mathcal{A}$-module, does it follow that they generate $\mathcal{F}(X)$ as an $\mathcal{A}(X)$-module?
The answer is negative and I will proceed to describe a counterexample from basic algebraic geometry. I will try to describe it first assuming no knowledge of algebraic geometry, and then make a few comments that assume some knowledge of the topic.
Let $k$ be a field that, for the sake of presentational simplicity, I will take to be algebraically closed. As a topological space, we consider $X := \mathbb{P}^1(k)$ (this is just $k \cup \{\infty\}$ if you will) with the Zariski topology, meaning that the open sets are the empty set and the cofinite sets. Call $(x{:}y)$ the homogeneous coordinates on $\mathbb{P}^1$, meaning that $(x{:}y)$ is the point $x/y$ except when $y=0$ and $x\neq 0$ in which case it is $\infty$ (and $(0{:}0)$ is not defined).
For $\ell\in\mathbb{Z}$, we define the sheaf $\mathcal{O}(\ell)$ on $\mathbb{P}^1$ as follows: if $U = X \setminus \{p_1,\ldots,p_k\}$ then the sections of $\mathcal{O}(\ell)$ on $U$ are homogeneous rational functions of total degree $\ell$ in $x,y$ (meaning the ratio of two homogeneous polynomials such as the degree of the numerator minus the degree of the denominator equals $\ell$) whose reduced denominator does not vanish except possibly at $p_1,\ldots,p_k$ (to be completely clear: for $q\neq\infty$, not vanishing at $q$ means that if you put $y=1$ and $x=q$ it does not vanish, and for $q=\infty$ it means that if you put $x=1$ and $y=0$ it does not vanish).
One can check that $\mathcal{A} := \mathcal{O} := \mathcal{O}(0)$ is a sheaf of rings on $\mathbb{P}^1$, and $\mathcal{O}(\ell)$ are sheaves of $\mathcal{O}$-modules.
Now choose a homogeneous polynomial $f$ of degree $2$ in $x,y$, and consider the $\mathcal{O}$-linear map$$\mathcal{O}(-2) \xrightarrow{f} \mathcal{O}$$taking $h$ to the product $f\cdot h$ (on any open set). This is obviously injective: let $\mathcal{F}$ be the cokernel, i.o.w., we have a short exact sequence$$0 \to \mathcal{O}(-2) \xrightarrow{f} \mathcal{O} \to \mathcal{F} \to 0 \tag{*}\label{02c81c59-fba9-478b-9d5b-e3ee95af1ce3}$$of $\mathcal{O}$-modules.
By definition, $\mathcal{F}$ is generated by a single global section, viꝫ. the image of the (unique up to multiplication by a constant in $k$) global section $1$ of $\mathcal{O}$. However, $\mathcal{O}(X) \to \mathcal{F}(X)$ is not surjective: to simplify the computational verification and since I just need one counterexample, let me show this if I take $f := x^2$. In this case, we define a section of $\mathcal{F}$ not proportional to (the image of) $1$ by taking the image of $\frac{x}{y} \in \mathcal{O}(U_1)$ over $U_1 := X \setminus\{\infty\}$ and the image of $0 \in \mathcal{O}(U_2)$ over $U_2 := X \setminus\{0\}$: since on $U_1\cap U_2 = X \setminus \{0,\infty\}$ we have $\frac{x}{y} = x^2 \cdot \frac{1}{xy}$ these two sections glue to form a section of $\mathcal{F}$ on $U_1 \cup U_2 = X$, which is clearly not proportional to $1$. So $\mathcal{F}(X)$ has dimension at least $2$ as a $k$-vector space (actually it's exactly $2$), whereas $\mathcal{O}(X)$ has dimension $1$. So $\mathcal{O} \to \mathcal{F}$ is surjective but $\mathcal{O}(X) \to \mathcal{F}(X)$ is not.
(This concludes the elementary presentation of the counterexample.)
The more sophisticated way to see this counterexample is, of course, to introduce cohomology groups: from $\eqref{02c81c59-fba9-478b-9d5b-e3ee95af1ce3}$ we get a long exact sequence in cohomology,$$\begin{aligned}0 \to& H^0(\mathbb{P}^1,\mathcal{O}(-2)) \to H^0(\mathbb{P}^1,\mathcal{O}) \to H^0(\mathbb{P}^1,\mathcal{F})\\\to& H^1(\mathbb{P}^1,\mathcal{O}(-2)) \to H^1(\mathbb{P}^1,\mathcal{O})\end{aligned}$$and as it is known from basic results in coherent sheaf cohomology that $H^0(\mathbb{P}^1,\mathcal{O}(\ell))$ has dimension $\max(0,\,\ell+1)$ and $H^1(\mathbb{P}^1,\mathcal{O}(\ell))$ has dimension $\max(0,\,-\ell-1)$ it follows that the map $H^0(\mathbb{P}^1,\mathcal{O}) \to H^0(\mathbb{P}^1,\mathcal{F})$ (i.e., $\mathcal{O}(X) \to \mathcal{F}(X)$) has cokernel of dimension $1$.
It is probably also worth pointing out that the image of $\mathcal{O}(-2) \xrightarrow{f} \mathcal{O}$ is the ideal sheaf generated by $f$ and the cokernel $\mathcal{F}$ is (the pushforward to $\mathbb{P}^1$ of) the structure sheaf $\mathcal{O}_Z$ of the zero-dimensional subscheme $Z := \{f=0\}$. In other words, we are dealing with the short exact sequence$$0 \to \mathcal{I}_Z \to \mathcal{O} \to \mathcal{O}_Z \to 0$$which says that locally a regular function on $Z$ is just the restriction of a regular function on $\mathbb{P}^1$, but evidently this is not true globally because the only regular functions on $\mathbb{P}^1$ are the constant ones while the fact that $Z$ has length $2$ means that (regardless of whether it is two distinct points or one thickened point) there are non-constant functions on it.