Switch – Store and forward latency

The switch will fully load the incoming frames of data, from the two sending systems, into its buffer(s). I'm not sure how it determines which frame would be first in the queue for subsequent forwarding; but it's probably based on initial receive time of the beginning of the frame. Then the switch works through the transmit buffer queue sending the frames out one-by-one onto the destination port/segment.

There's no issue with frames "running into each other." The real issue is can the ultimate port/segment accept the frames fast enough. (And, of course, can the switch process its buffer/queues fast enough.)

It depends on several factors.

IF there is only a single ring (eg. each switch is connected to exactly two switches using exactly one cable to each switch) and the source is connected to one of the switches then if you do a broadcast storm (destination MAC FF:FF:FF:FF:FF:FF) and if you would constantly be flooding frames from the source then it would utilize all of the links to the full capacity in both directions (it's full duplex).

If you would for example only send one frame and let it circulate, the utilization would depend on how quickly a switch can forward that frame. Then you'd need to sum up the forwarding, propagation, serialization and other delays in the entire loop and that often would that one (actually two, one in each direction) frame arrive on each particular interface. Then based on the frame size you would simply calculate the utilization.

But remember, utilization is always a function of time, there is no such thing as instant utilization.

So to answer the formula question: utilization of each single link in one particular direction = (num_frames * frame_size / sum_of_all_delays_in_the_path) / capacity_of_the_link