For a particle upon which a force field $\vec{F}$ is applied, its trajectory $\mathcal{C}$ (if the problem is studied in two dimensions) can be parametrized by a real parameter $t$ (time), so it is given by $\vec{r}(t)=(x(t),y(t))$ with $t \in [t_i,t_f]$.
...despite the fact of there existing a force field (in terms of position) which may contradict the path in terms of time.
Even though you parametrize the path, in order to evaluate the line integral you must apply this parametrization to both components of the vector field, so you get the components of the force field in terms of time:
\begin{align}
F_x&=P(x(t),y(t)) \\
F_y&=Q(x(t),y(t)).
\end{align}
This way, the line integral is given by
\begin{align}
\int_{\mathcal{C}} \vec{F}\cdot d\vec{r} &= \int_{\mathcal{C}} P(x,y)dx+Q(x,y)dy \\
&= \int^{t_f}_{t_i} P(x(t),y(t))\frac{dx(t)}{dt}dt + \int^{t_f}_{t_i} Q(x(t),y(t))\frac{dy(t)}{dt}dt.
\end{align}
...how could I model the particle's path in terms of time (i.e. a parametric path)?
There is not a unique way to parametrize the path taken by the particle, but for simple trajectories like a circumference or straight lines, it's quite easy. For example, suppose the particle travels about a semi-circunference of radius R anti-clockwise, going from $(x,y)=(R,0)$ to $(x,y)=(-R,0)$. Then, a valid parametrization of $\vec{r}$ is the following:
\begin{align}
x(t)&=R\cos(t) \\
y(t)&=R\sin(t),
\end{align}
with $t \in [0,\pi]$ (you can evaluate $t=0$ and $t=\pi$ into the parametrization to make sure that this interval in effect represent the path given). Applying this parametrization to the components of the vector field involved in the problem and differentiating $x(t)$ and $y(t)$ with respect to the parameter, you can evaluate the integral given above.
