It always helps to sketch the support of the joint distribution; i.e., the region of $(X,Y)$ such that the density is positive. In your case, you can see that this region must be in the upper half plane ($y \ge 0$), and for a given value of $y$, $x$ must be between $-y$ and $y$; so this region is triangular and is bounded to the right by $y = x$ and to the left by $y = -x$.
Thus, the marginal density of $X$ is given by the integral $$f_X(x) = \int_{y=|x|}^\infty f_{X,Y}(x,y) \, dy$$ since we require $y$ to be at least as large as $|x|$ (otherwise the joint density is zero). This corresponds to picking an $x$-value, drawing a vertical line, and integrating over a $y$-interval that corresponds to the intersection of this vertical line and the support of the joint distribution; in a sense, it is like "collapsing" the support onto the $x$-axis, where the marginal density at each $x$-value is the integral of the joint density for all corresponding $y$-values.
The marginal density of $Y$ is easier to see: $$f_Y(y) = \int_{x=-y}^y f_{X,Y}(x,y) \, dx$$ as you wrote. This is analogous to "collapsing" the support onto the $y$-axis, "summing" up the joint density in horizontal rather than vertical lines.