Suppose that $\phi$ is a $1-cycle$ in a topological space $X$, and that $\phi^*:= \phi r$ is its opposite orientation where $r(t,1-t)=r(1-t,t)$. How should one show that there is a $2-chain$ whose boundary is $\phi+\phi^*$?
I know a proof for a more general fact (given any 1-simplex, the sum of it and its opposite orientation is a 1-boundary), but for this simple case is there any simple solution?
I am going to assume that you are asking about the singular chain complex of $X$, so in this answer a "simplex" means a "singular simplex in the topological space $X$".
Suppose at first that one knows how to find a solution for each 1-simplex $\sigma$, i.e. one knows how to find $\tau$ so that $\partial\tau = \sigma + \sigma^*$.
Consider any $1$-chain $\phi$ (not just a 1-cycle). We may write $\phi$ as a finite linear combination of 1-simplices, $$\phi = \sum_i a_i \sigma_i $$ where each $\sigma_i$ is a 1-simplex and each $a_i$ is a coefficient. Using your proof for a 1-simplex, it follows that for each $i$ there is a 2-chain $\tau_i$ such that $\partial \tau_i = \sigma_i + \sigma^*_i$. From this we obtain $$\phi + \phi^* = \sum_i a_i \sigma_i + \sum_i a_i \sigma^*_i = \sum_i a_i (\sigma_i + \sigma^*_i) = \partial \left( \sum_i a_i \tau_i \right) $$ and so $\sum_i a_i \tau_i$ is the required 2-chain.
So to complete the answer, I'll describe $\tau$ in the special case of a single 1-simplex $\sigma$ (which does not have to be a cycle). I'll follow the standard notations in many algebraic topology books, for example Hatcher's book. By definition, $\sigma$ is a continuous function $\sigma : \Delta^1 \to X$. Let $c : \Delta^2 \to \Delta^1$ be the affine map that maps the first and third vertices of $\Delta^2$ to the first vertex of $\Delta^1$, and maps the second vertex of $\Delta^2$ to the second vertex of $\Delta^1$. Let $d : \Delta^2 \to \Delta^1$ be the constant map which take $\Delta^2$ to the first vertex of $\Delta^1$. Then we just take $\tau = \sigma \circ c + \sigma \circ d$. A short computation shows that $\partial \tau = \sigma + \sigma^*$.