How can sum of component forces of gravity on inclined plane be greater than actual force of gravity?

You're adding the magnitudes of the component forces. This gives an answer that is too large, because the component forces aren't pulling in the same direction as each other. The component forces, $vec{A}$ and $vec{B}$, say, need to be added as vectors, that is represented as arrows of length proportional to the magnitude of the components and pointing in the direction of these components. Arrange them tail of $vec{B}$ to head of $vec{A}$ and the resultant force (the vector sum of $vec{A}$ and $vec{B}$) os represented by an arrow going from the tail of $vec{A}$ to head of $vec{B}$.

You can do this using a diagram drawn to scale, or a sketched diagram to which you apply trigonometry. When you've done this sort of thing two or three times you don't even need a sketched vector diagram!

A couple of additional remarks that may (or may not) help…

(1) You've resolved the weight into a force of $500 frac{1}{sqrt{2}}$ N normally into the slope, and a force of $500 frac{1}{sqrt{2}}$ N 'down' along the slope. While the downward components (each of $500 frac{1}{sqrt{2}} frac{1}{sqrt{2}}$ N) of these two component forces add up as scalars to make the original downward force, their horizontal components, each of $500 frac{1}{sqrt{2}} frac{1}{sqrt{2}}$ N but in opposite directions, cancel.

(2) Force is one of a 'family' of Euclidian vectors. The archetypal Euclidian vector is displacement. 500 m North East is an example of a displacement. Suppose that we wanted to get from A to B, a displacement of 500 m North East. Suppose also that roads and rail were only available in a grid pattern, going East-West (and West-East) and North South (and South North). To get from A to B by road or rail you'd have to go $500 frac{1}{sqrt{2}}$ m North and $500 frac{1}{sqrt{2}}$ m East. That means you've gone a total distance of $500 sqrt{2}$ m. We've just added the magnitudes (scalars) of the two component vectors, and found a figure that is greater than the magnitude (500 m) of our displacement. But the two component displacements ($500 frac{1}{sqrt{2}}$ m North and $500 frac{1}{sqrt{2}}$ m East) add as vectors to give the required displacement of 500 m North East!