The force (on a single point particle) is also simply a function:
$$\begin{cases}
\mathbf F: \mathbb R \rightarrow \mathbb R^3\\
t \longmapsto \mathbf F(t)
\end{cases}$$
In other words, $\mathbf F$ takes time as an input and spits out the force exerted on the particle at that time as its output. <br><br>

However, I assume your confusion comes from the definition of work, usually stated as:
$$W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$$
where $C_{AB}$ is the path that the particle takes while going from point $A$ to point $B$. However, with a bit of nitpicking, this notation may be somewhat misleading. I would choose to write the definition of work in this way instead:
$$W_{A\rightarrow B} \equiv \int_{t_A}^{t_B}d\mathbf r(t) \cdot \mathbf F(t) \equiv \int_{t_A}^{t_B}dt \ \mathbf v(t) \cdot \mathbf F(t) \qquad (*)$$
where I used $d\mathbf r(t) = \frac{d\mathbf r}{dt} \ dt = \mathbf v(t) dt$. If you're wondering why people sometimes write $\mathbf F(\mathbf r)$, see the last part of my answer. <br>

This implies that as a mathematical object, work can be thought of as a *functional*, which takes the two functions $\mathbf v: \mathbb R \rightarrow \mathbb R^3$ and $\mathbf F: \mathbb R \rightarrow \mathbb R^3$, and returns a real number. So if I call the set of "sufficiently well-behaved" (sorry this is as rigorous as I can get ;)) functions from $\mathbb R$ to $\mathbb R^3$ simply $\mathfrak F$, the work functional can be characterized as:
$$\begin{cases}
W: \mathfrak F \times \mathfrak F \rightarrow \mathbb R\\
(\mathbf v, \mathbf F) \longmapsto W[\mathbf v,\mathbf F]
\end{cases}$$
In other words, the work functional takes the velocity function, or equivalently the position (disregarding a constant shift), as well as the force function, and spits out a number.


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You might be asking why the force is usually written as $\mathbf F(\mathbf r)$, instead of $\mathbf F(t)$.This is because most of the time, people are thinking of a *force field*, which is technically a different mathematical object:
$$\begin{cases}
\mathbf f: \mathbb R^3 \rightarrow \mathbb R^3\\
\mathbf r \longmapsto \mathbf f(\mathbf r)
\end{cases}$$
A force field takes a position vector $\mathbf r$, and spits out the force $\mathbf f$ exerted on a particle (with a specific mass, etc.) *if it was sitting at* $\mathbf r$. For example, a gravitational force field $\mathbf f(\mathbf r) = GMm \ {\mathbf r}/{|\mathbf r|^3}$ tells you that if you had a particle of mass $m$ at position $\mathbf r$, the force on that particle would be $\mathbf f(\mathbf r)$. <br>

Now the force $\mathbf F: \mathbb R \rightarrow \mathbb R^3$ in the case of a force field is simply a composition of the two functions $\mathbf f: \mathbb R^3 \rightarrow \mathbb R^3$ and $\mathbf r: \mathbb R \rightarrow \mathbb R^3$. In other words, the force at time $t$ is simply the force field evaluated at the position at which the particle is at that instant. So:
$$\mathbf F(t) \equiv \mathbf f(\mathbf r(t))$$
Using this on the definition of work $(*)$ you get:
$$W_{A \rightarrow B} = \int_{t_A}^{t_B} d\mathbf r(t) \cdot \mathbf f(\mathbf r(t))$$
which is what people actually mean by $W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$.