If $g$ is a (not necessarily closed) path in X, prove that the $1$-chain $g$ is homologous to $-g^{-1}$.
What exactly does it mean for these two $1$-chains to be homologous? Does it mean $\text{ cls } g = \text{ cls } (-g^{-1})$? And if so what exactly is a negative chain imply?
When we say that $c$ is homologous to $c'$, we usually mean they represent the same element in the homology group, that is, $[c] = [c']$, as you expected. This, however, only makes sense when $c$ and $c'$ are cycles. if they are not, we can still call them homologous if $$ c - c' = \partial b $$ for some $2$-chain $b$. Note that this means $c-c'$ is a cycle, so only $1$-chains with the same boundary can be homologous.
An example of a situation where this extended definition of homologous is opportune is when we compare the fundamental group of a based space $(X,x_0)$ with the homology group $H_1(X)$. Here homotopic pathes $f$ and $g$ give rise to homologous simplices $f,g: \Delta^1 \to X$. We therefore have a map $\pi_1(X,x_0) \to H_1(X)$.