Instead of cutting corners and jumping right in to a reduction, I believe you would profit from actually stepping back and clarifying the logical basis of your argument.
Given an instance $(G, s, t)$ of HAMPATH, your reduction produces an instance $(G', s')$ of HAMCYCLE, as you have described. To prove your reduction works, you need to show:
$$(G, s, t) \in \textsf{HAMPATH} \iff (G', s') \in \textsf{HAMCYCLE}$$
for every instance $(G, s, t)$ of HAMPATH, where $(G', s')$ is uniquely determined as a function of $(G, s, t)$ (the function in case being your reduction).
To prove this equivalence, you first fix an arbitrary instance $(G, s, t)$ of HAMPATH (which may be a "yes" or a "no" instance). $(G', s')$ is then uniquely determined by $(G, s, t)$. Then, you either prove the two sides of the equivalence (as you have done) or show that, if $(G, s, t)$ is a "yes" instance, then $(G', s')$ is a "yes" instance and, if $(G, s, t)$ is a "no" instance, then $(G', s')$ is a "no" instance. Although usually the latter is preferred, either is fine; you can reference all of $G$, $s$, $t$, $G'$, and $s'$ in both cases since they are all fixed.
The problem with your argument is that you obtain a cycle $(s, \dots, s)$ and then (correctly) reason there is a node $t'$ right before $s$ which makes it a Hamiltonian cycle (and gives you a Hamiltonian path from $s$ to $t'$). However, because $t$ is fixed, there is no guarantee that $t' = t$. You still need to consider the case $t' \neq t$ and show this gives you a Hamiltonian path from $s$ to $t$ even if you delete the edge $(t, s)$ (since this yields $G$ from $G'$).
Unfortunately, you will not be able to. Consider, for instance, the case where $G$ is a cycle $(s, t, v, s)$. Then both $G$ and $G'$ obviously have a Hamiltonian cycle which starts and ends at $s$, but there is no Hamiltonian path from $s$ to $t$ in $G$; the only way to reach $t$ from $s$ (by a path) is the single edge $(s, t)$, which fails to visit the vertex $v$.
You will have to modify your reduction. Hint: Try to find a way of forcing $G$ to not have a Hamiltonian cycle in the first place (and note this implies $t' = t$ above).
