This is the core idea of Bayesian statistics.
In a nutshell, you start with some prior beliefs. This might be a simple non-informative prior ('the teams are equally likely to win the series'), or it could be the result of a complex model that includes things like their prior match-ups, when and where they're playing etc.
When you receive new information, you combine the prior with that information to generate a posterior probability. This is done using a likelihood function. You do this over and over, using the posterior from one round as the prior for the next. The nice thing about this is that posterior represents your (well, the model's) beliefs at any point in time.
In general, the math for this can be hard-to-totally-infeasible, because you're multiplying, normalizing, and marginalizing arbitrary probability distributions. However there are some tricks can dramatically simplify it. One is to use a conjugate prior. This essentially boils down to choosing ways to represent your data so that the prior and posterior have the same form. For example, here's a worked example. Alternately, you can sidestep the difficulties involved in finding a closed-form solution and use something like Markov Chain Monte Carlo, which approximates the probability distribution numerically (tutorial).
Edit: To your specific question: yes, you almost have to update the probabilities as you go. The goal of Bayesian statistics is to find probabilities that accurately reflect your beliefs. If a team is down by 3 games and you believe that there is only a 60% chance that they'll win the next game (and avoid elimination), you certainly shouldn't report a 70% chance that they'll win the whole series!
The approach you described above seems totally reasonable--you'd just recalculate the probability of winning $m$ of the next $n$ games given your assumed $P(\textrm{win})=0.6$ and with $m$ and $n$ chosen based on the outcomes so far. For example, you'd need the probability of four consecutive wins for a team that's down 0-3 to win a best-of-seven series, which is $(0.6)^4\approx 0.13$ A more sophisticated analysis might also consider that the "true" (but unknowable) property of the team winning might not be 60 percent, and you might also want to update that too.
