Without more detailed information about the relationship between the metered time and the correction in stops, I’d construct a simple linear model for each time interval in the table. They should give a reasonable estimate of the exposure correction required.
Using (t, Δ) to represent a point containing time t and Δ to represent the difference in stops, we have points (1 second, 1/3 stop) and (10 seconds, 1/2 stop). Using linear regression, we get
Δ(t) = 0.0185185t + 0.3148148 for t within the interval from 1 to 10 seconds
At t = 1 sec
Δ(1 sec) = 0.0185185(1 sec) + 0.31414 = 1/3 stop
and at 10 seconds,
Δ(10 sec) = 0.0185185(1 sec) + 0.31414 = 1/2 stop
So, for a time of t = 6 seconds, we get
Δ(6 sec) = 0.0185185(6 sec) + 0.31414 = 0.4259 stop
For metered times from 10 seconds to 100 seconds,
Δ(t) = 0.0055555t + 0.44444
Δ(10 sec) = 0.0055555(10 sec) + 0.44444 = 1/2 stop and
Δ(100 sec) = 0.0055555(100 sec) + 0.44444 = 1 stop
At 60 seconds, the model predicts
Δ(60 sec) = 0.0055555(60 sec) + 0.44444 = 0.78 stop
As noted in post #2, since this is negative film and the time intervals are relatively small, the exposure correction isn’t overly fussy. For most uses, the model in this post (post #3) is overkill, but it can be done if wanted.