Solving $UCYCLE$ in logspace - two possible approaches ? Why can't we one of them use to solve connectivity?

$$UCYLE = \mathcal \{ <G> ~:~ G \text{ is an undirected graph that contains a simple cycle}\}.$$
There are two possible approaches to this exercise: Solving cycle in undirected graph in log space?

These two solutions are different. The instresting solution is second one (on stackoverflow). In this case we generally implement DFS in logspace. I can see that it does work (using DFS). However the question is:
Why we can't use this DFS algorithm to solve USTCON problem ?

However, I can see that we can't easily use this DFS to solve bipartity problem. It is not easy to count number of nodes in cycle.

Edit
Modified algorithm from stackoverflow (mentioned link) which should solve ustcon problem:

for v_j in neighbours(s)
current, prev = v_j, s
repeat
idx = neighbours(current).index(v_j)
idx = (idx + 1) % len(neighbours(current))
current, prev = neighbours(current)[idx], current
if current = t then return connection between s and t
until current adjacent to s
return no_connection between s and t
• Why do you think we can't use it? What we can't is to show that USTCON is in $\mathsf{L}$ using it. Because you will need to keep in memory every adjacent cycle. – rus9384 Aug 9 '17 at 18:12
• @rus9384 I don't understand you. This algorithm is simply DFS in logspace. So we should be able to check connectivity between two verticles using this DFS. – Haskell Fun Aug 9 '17 at 18:22
• Strongly connected component is not necessarily a cycle, it can be several adjacent cycles. You need to use contraction in that case which is not in log space. – rus9384 Aug 9 '17 at 18:57
• Strongly connected component - it is about directed graphs, I consider only undirected. Once again - we have DFS in logspace. It allows us to check connectivity between two verticles in logspace. – Haskell Fun Aug 9 '17 at 19:05
• Hm, in fact, thought are these: DFS algorithm only counts cycles where each vertex appear once. However, such cycles that are constructed of smaller cycles, as 8 for example, cannot be found using that algorithm. – rus9384 Aug 9 '17 at 19:15