forecast() function details

I have a code which takes the input as the Yield Spread (dependent var.) and Forward Rates(independent var.) and operate an auto.arima to get the orders. Afterwards, I am forecasting the next 25 dates (forc.horizon). My training data are the first 600 (training). Then I am moving the time window 25 dates, meaning using the data from 26 to 625, estimating the auto.arima and then forecasting the data from 626 to 650 and so on. My data sets are 2298 rows (date) and 30 columns (maturity).

I want to store all of the forecasts and then plot the forecasted and real values in the same plot.

My main question is how forecast function does forecasting? For example if I give xreg 25 new data, does it take the 601 value and does forecast for 601th dependent variable or does it take all 25 into account and does forecasting?
My main concern is that it would use future information to forecast today's value.
Here is the like to my data, it is formatted in .rda so you can easily open it in R studio:

Here's my code:


forecast.func <- function(NS.spread, ind.v, maturity, training, forc.horizon){
  NS.spread <- NS.spread/100
  forc <- c()
  j <- 0
  for(i in 1:floor((nrow(NS.spread)-training)/forc.horizon)){
    # linear model
    y <- as.vector(NS.spread[(1+j):(training+j) , maturity])
    f <- ind.v[(1+j):(training+j) , maturity]
    # auto- arima
    c <- auto.arima(y, xreg = f, test= "adf")

    # forecast
    e <- ind.v[(training+j+1):(training+j+forc.horizon) , maturity]
    h <- forecast(c, xreg = lagmatrix(e, -1))
    forc <- cbind(forc, h)
    j <- j + forc.horizon


a <- forecast.func(spread.NS.JPM, Forward.rate.JPM, 10, 600, 25)
lapply(a$x, plot)
lapply(a$fitted, lines, col = "red")

You should check the package author's online text
and lecture notes

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