What you were told is not accurate, or rather, technically accurate if you consider it as abuse of notation, but very much unhelpful, and not a definition by any means. In the context of physics, a curve $C$ is merely a function $C:[t_0,t_1]\rightarrow\mathbb{R}^n$ satisfying some differentiability requirements. If $\mathbf{F}$ is some vector field, then $$\int_C\mathbf{F}(\mathbf{r})\cdot{\mathrm{d}\mathbf{r}}:=\int_{t_0}^{t_1}\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t)\,\mathrm{d}t$$ which is well defined, since $$\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t)$$ is just some $f(t)$ for some function $f$, because dot products always result in scalar quantities. This is indeed, the correct definition. However, you can do some things here to arrive at the result that you were presented with. If $n=2$, and you are given an orthonormal basis $B=\{\mathbf{e_x},\mathbf{e_y}\}$, then $$\mathbf{F}=F_x\mathbf{e_x}+F_y\mathbf{e_y}=(F_x,F_y)_B$$ and $$\mathbf{r'}(t)=(x'(t),y'(t))_B,$$ implying $$\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t)=(F_x(\mathbf{r}(t)),F_y(\mathbf{r}(t)))_B\cdot(x'(t),y'(t))_B.$$ Now, in an abuse of notation, one may consider $\mathrm{d}x=x'(t)\,\mathrm{d}t$ and $\mathrm{d}y=y'(t)\,\mathrm{d}t$. All in all, one would have $$\int_{t_0}^{t_1}\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t)\,\mathrm{d}t=\int_{t_0}^{t_1}(F_x(\mathbf{r}(t)),F_y(\mathbf{r}(t)))_B\cdot(x'(t),y'(t))_B\,\mathrm{d}t=\int_C(F_x(\mathbf{r}),F_y(\mathbf{r}))_B\cdot(\mathrm{d}x,\mathrm{d}y)_B,$$ which in your post, is more shortly denoted as $$\int_C(F_x,F_y)\cdot(\mathrm{d}x,\mathrm{d}y),$$ not specifying the basis. And again, this is abuse of notation, because it treats the symbol $\mathrm{d}t$ in the notation of the Riemann integral as an actual mathematical object with its own objective, independent existence, which is not the case at all. And this all assumes a specific type of basis, which is inadequate, because in physics, these path-integrals are basis-independent, which is to say, they form tensors: they are invariants. So a specific basis for the dot product should not be chosen. It also assumes specifically that space is 2-dimensional. This last one is not a particularly big deal, but it still makes the formula you were presented with very awkward, since what you were presented with is presumably supposed to be a definition, not a circumstance-based theorem.
Hopefully this clarifies any doubts you have, and more importantly, you hopefully have a better understanding of what a path/line integral of a vector field actually is.
Best Answer
Suppose ${\bf F}=(f(x,y),g(x,y)$ and then $$ \sqrt{f^2(x,y)+g^2(x,y)}\le M. $$ Let ${\bf r}=(x(t),y(t))$ be the parametric expression of $C$ for $t\in[0,l]$. Then $$ d{\bf r}=(x'(t),y'(t))dt $$ and hence $$ \int_C{\bf{F}}\cdot d{\bf r}=\int_0^l(f(x,y)x'+g(x,y)y')dt. $$ Using $$ |a_1b_1+a_2b_2|\le \sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2} $$ one has \begin{eqnarray} &&\bigg|\int_C{\bf{F}}\cdot d{\bf r}\bigg|\\ &=&\bigg|\int_0^l(f(x,y)x'+g(x,y)y')dt\bigg|\\ &\le&\int_0^l\bigg|f(x,y)x'+g(x,y)y')\bigg|dt\\ &\le&\int_0^l\sqrt{f^2(x,y)+g^2(x,y)}\sqrt{(x')^2+(y')^2}dt\\ &\le&M\int_0^l\sqrt{(x')^2+(y')^2}dt\\ &=&ML. \end{eqnarray} Another way to do is to simply use $$ \bigg|\int_C{\bf{F}}\cdot d{\bf r}\bigg|\le \int_C\bigg|{\bf{F}}\cdot d{\bf r}\bigg|\le \int_C|{\bf{F}}||d{\bf r}|\le M\int_C|d{\bf r}|=ML. $$