The Kaplan-Meier estimator for the survival function and the Nelson-Aalen estimator for the cumulative hazard have step changes at event times, so trying to differentiate with respect to time (as you suggested in a comment) to get the instantaneous hazard won't work. The derivative will be 0 between event times and infinite at event times.
It's not clear why you need to know the instantaneous hazard as a function of time if your interest is in a cumulative hazard like your $F(t)$. If there are no covariates to account for, then it seems that what you want is the Nelson-Aalen estimate itself, which describes cumulative hazard for the data set at hand while taking censoring into account. If you wish to control for covariates, then you could fit, say, a Cox proportional hazards model with a statistical software package and use the associated prediction functions to generate plots that illustrate how those covariates affect survival. The survival functions will still have step changes at event times, however.
If you want a smooth approximation to the data then you should fit some parametric survival model to the data, again potentially including covariates. A Weibull model often works well and has more flexibility than the constant-hazard model implicit in your equation for $F(t)$; that constant-hazard model is one particular example of the more general Weibull model. That's one of several parametric possibilities provided by standard statistical software packages. With such a smooth parametric approximation to the survival curve then you can differentiate with respect to time and get instantaneous hazards if you still need that.
In response to information added to question:
Your data are certainly in a form suitable for standard survival analysis (Kaplan-Meier, Cox models, parametric survival regression): a wide enough range of times (from 6 to 60) that it doesn't seem necessary to use discrete-time analysis, some covariates to consider, and annotations of event (default) versus censored (no default at last observation time), with a separate row containing these values for each individual. For predictions about survival based on your data, you don't have to re-invent the methods yourself (as the start of your question might be taken to mean). You can just use the modeling and prediction facilities provided by any standard statistical software system that handles survival data, which incorporate all the issues that you raise (and more).
For example, in the basic R
survival package there is a
predict() function that can handle either semi-parametric Cox proportional hazards models (
coxph() with its implicit baseline hazards) or fully parametric models (
survreg() with 6 built-in choices of assumed distributions, including Weibull). You fit a model to the data you have, then provide the model along with the covariate values for any new case for which you wish a survival prediction. (It's good practice also to request the standard errors for the prediction.)
The above assumes, however, that
DaysSurvived=0 in your data represents some appropriate reference time that applies equally for all cases, for example the date on which the loan was made to that individual, and that the covariate values are those that held for each individual at the corresponding
DaysSurvived=0. Otherwise a more complicated analysis might be necessary.