Some good answers have already been given. To my mind, though: a really obvious thing one wants to do in topology is to classify topological spaces, or more reasonably, to classify CW-complexes, at least up to homotopy equivalence.

• You build CW-complexes by attaching cells.

• To attach a cell, you just have to specify the attaching map, i.e., the continuous map from the boundary of the cell to the skeleton you've already constructed.

• The boundary of a cell is a sphere.

• The homotopy type of the space you get by attaching a cell depends only on the homotopy class of the attaching map.

So homotopy groups are telling you the ways you can attach a cell. So the question of how many homotopy types of CW-complexes you can build with some given property typically comes down to a computational problem about homotopy groups.

For example: it sounds like you are an arithmetic geometer who already is convinced of the utility of homology groups and the fundamental groups, so suppose you are doing your work in arithmetic geometry, and you find yourself confronted with two different smooth schemes over Spec Z whose underlying analytic spaces are simply-connected and have homology groups isomorphic to Z in degrees 0, 8, and 14, and trivial in all other degrees. You knock on the door of the friendly topologist down the hall and ask the topologist whether your two analytic spaces are necessarily homotopy-equivalent.

• The topologist first checks with you to make sure there's some general theorem which ensures that these spaces are homotopy-equivalent to CW-complexes, and then observes that a minimal CW-decomposition of any such space ought to have a 0-cell, an 8-cell, and an 14-cell, since anything else would give you the wrong homology.
• The 8-cell has to be attached trivially to the 0-cell for silly reasons, so the 8-skeleton of any such CW-complex must be S^8.
• Then the topologist points out that the attaching map for the 14-cell must be a map from its boundary, a 13-sphere, to an 8-sphere.
• The homotopy group $\pi_{13}(S^8)$ is in the stable range, by the Freudenthal suspension theorem, so the topologist shows you a 2-primary Adams spectral sequence chart and points to the empty 5-column, and says "So the 5-stem is 2-locally trivial."
• Then the topologist tells you a bit (maybe more than you wanted to hear) about how the alpha family and beta_1 work at odd primes, ending with the conclusion that the 5-stem (the fifth stable homotopy group of spheres) also vanishes at all odd primes, and hence $\pi_{13}(S^8)$ is trivial.
• Consequently there is only one homotopy class of attaching map for that 14-cell which has been attached to S^8. So your two analytic spaces are homotopy-equivalent.

If you have more than just two cells in positive dimensions, Toda brackets are a convenient way to organize the algebra of the attaching maps, in order to reduce these kinds of classification problems to algebraic problems in the homotopy groups of spheres.

Never used this site before--hope I didn't write anything too critically stupid. Sorry if I did.