You need to add the word "oriented", in order to assign signs consistently to the intersections, and "compact" so that there will be finitely many intersections. To see the issue for unoriented manifolds, think about $\mathbb{RP}^2$. We have $H_0(\mathbb{RP}^2) \cong \mathbb{Z}$ and $H_1(\mathbb{RP}^2) \cong \mathbb{Z}/2$, so cap product would define a bilinear map $\mathbb{Z}/2 \times \mathbb{Z}/2 \to \mathbb{Z}$. Such a map must be zero, but two non-contractible curves in $\mathbb{RP}^2$ meet an odd number of times. (For example, two lines in $\mathbb{RP}^2$ meet at a single point.)

After that, nothing is wrong with it. For a compact oriented manifold $X$, Poincare duality gives an isomorphism $H^k(X) \cong H_{n-k}(X)$ so we get a map
$$H_{n-a} \times H_{n-b} \cong H^a \times H^b \overset{\cup}{\longrightarrow} H^{a+b} \cong H_{n-a-b}.$$
Tracing through the definitions on a triangulation, one obtains the map you described.

As far as I know, it is only important to distinguish cap and cup when one wants to move away from the world of compact oriented manifolds. (Cup makes sense when $X$ isn't a manifold at all!)