From Rotman's Algebraic Topology concerning developing the intuition on homology functors:
The question we ask is whether a union of $n$-smplexes in $X$ actually is such a boundary. Consider the case $n=1.$ Let $X$ be the punctured plane $\Bbb R^2 - \{0\}$, and let $\alpha, \beta, \gamma$ be the $1$-simplexes as drawn (the picture shows a triangle with the origin missing inside the triangle with edges $\alpha, \beta, \gamma$ oriented counter-clockwise); $\alpha \cup \beta \cup \gamma$ "ought" to bound the triangular $2$-simplex, but the absence of the origin prevents this; loosely speaking, $X$ has a "one-dimensional" hole in it. (Of course, $\alpha \cup \beta \cup \gamma$ would not bound the triangular $2$-simplex if $X$ were missing a small neighborhood of the origin or a small line segment through the origin).
How does $T = \alpha \cup \beta \cup \gamma$ NOT bound the triangular $2$-simplex? Isn't all of the triangular $2$-simplex within $T$? If you just fill in the inside of $T$ shouldn't you get the triangular $2$-simplex? And even if you take out lines through and neighborhoods of the origin doesn't this still bound the $2$-simplex?
Of course abstractly $T$ bounds the $2$-simplex $\Delta^2$, but not in $X$. There is no subspace of $X$ homeomorphic to $\Delta^2$ which is bounded by $T$.