What should I do to create new variables, using previous columns from my database

A fairly simple solution:

year <- c(1998, 1999, 2000, 2010)  
Age <- c(15, 16, 17, 18) 
'Birth Order 1' <- c(10, 25, 25, 35)
'Parity 0' <- c(100, 150, 140, 150)
'Birth Order 2' <- c(5, 10, 10, 30)
'Parity 1' <- c(110, 160, 150, 150)

mat <- data.frame(year, Age, `Birth Order 1`, `Birth Order 2`, `Parity 0`, `Parity 1`)

for(i in 1:2){
    mat[[paste0("mat",i)]] <- mat[[paste0("Birth.Order.",i)]]/mat[[paste0("Parity.",i-1)]]
}

mat

#  year Age Birth.Order.1 Birth.Order.2 Parity.0 Parity.1      mat1       mat2
#1 1998  15            10             5      100      110 0.1000000 0.04545455
#2 1999  16            25            10      150      160 0.1666667 0.06250000
#3 2000  17            25            10      140      150 0.1785714 0.06666667
#4 2010  18            35            30      150      150 0.2333333 0.20000000

You can use dplyr, combining arrange() and group_by() to then compute the lags of birth order / parity.

# toy data with 8 levels for birth order and parity
df <- data.frame(age=(15:18),year=c(1999:2002),
                 variable=c(
                   rep(paste0('birth.order',seq(1:8)),each=4),
                   rep(paste0('parity',seq(from=0,to=7,by=1)),each=4)),
                 value=c(
                   sample(10:40,32,T),
                   sample(100:150,32,T)))

# change to wide format, as in example
df <- df %>% spread(variable,value)

# gather, arrange, and compute lags, then output desired numbers
df %>% gather(k,v,-age,-year) %>% mutate(num=sub('.*([0-9])','\\1',k)) %>% 
  arrange(age,year,num,k) %>% group_by(age,year) %>% 
  mutate(mat=v/lag(v)) %>% filter(grepl('birth',k)) %>% 
  select(-v,-num)

# A tibble: 32 x 4
# Groups:   age, year [4]
     age  year k              mat
   <int> <int> <chr>        <dbl>
 1    15  1999 birth.order1 0.306
 2    15  1999 birth.order2 0.194
 3    15  1999 birth.order3 0.310
 4    15  1999 birth.order4 0.159
 5    15  1999 birth.order5 0.116
 6    15  1999 birth.order6 0.193
 7    15  1999 birth.order7 0.221
 8    15  1999 birth.order8 0.110
 9    16  2000 birth.order1 0.305
10    16  2000 birth.order2 0.264
# ... with 22 more rows

You can divide it directly in base R after identifying the columns

birth_cols <- grep("^Birth", names(mat))
parity_cols <- grep("^Parity", names(mat))
cbind(mat[1:2], mat[birth_cols]/mat[parity_cols])

#  year Age Birth.Order.1 Birth.Order.2
#1 1998  15     0.1000000    0.04545455
#2 1999  16     0.1666667    0.06250000
#3 2000  17     0.1785714    0.06666667
#4 2010  18     0.2333333    0.20000000

which is same as doing in dplyr as

library(dplyr)
mat %>%  select(starts_with("Birth")) / mat %>%  select(starts_with("Parity"))