这与OP的期望输出相匹配，并且很清楚每一步发生了什么。
首先，我们按school_id和subject对数据进行分组，以确定哪个原因是第一个，哪个原因是第二个，等等。然后我们基于school_id和subject透视表，使用Reason作为值，Reason_Name作为列名。

data %>% 
  group_by(school_id, subject) %>% 
  mutate(Reason_Name = paste0("Reason", row_number())) %>%
  ungroup() %>%
  pivot_wider(
    id_cols = c(school_id, subject, vacancies),
    values_from = Reason,
    names_from = Reason_Name
  ) 

# A tibble: 4 × 5
  school_id subject vacancies Reason1    Reason2 
  <chr>     <chr>       <dbl> <chr>      <chr>   
1 1         English        40 Pay        Location
2 2         Maths          20 Pay        NA      
3 3         English        10 Experience NA      
4 3         French         40 Pay        NA

如果我们想把这段代码压缩一点，我们可以说：

data %>% 
  mutate(
    Reason_Name = paste0("Reason", row_number()), 
    .by = c(school_id, subject))
  ) %>%
  pivot_wider(
    id_cols = c(school_id, subject, vacancies),
    values_from = Reason,
    names_from = Reason_Name
  )

library(tidyverse)
data %>%
   mutate(rowid = row_number(), .by = c(school_id, subject)) %>%
   pivot_wider(id_cols = c(school_id, subject, vacancies),
               names_from = rowid, 
               values_from = Reason,
               names_prefix = "Reason")

# A tibble: 4 × 5
  school_id subject vacancies Reason1    Reason2 
  <chr>     <chr>       <dbl> <chr>      <chr>   
1 1         English        40 Pay        Location
2 2         Maths          20 Pay        NA      
3 3         French         40 Experience NA      
4 3         French         10 NA         Location

这并不完全是你所描述的你想要的输出，但类似于GuedesBF的解决方案（我们有一个类似的思考过程，它出现;我是独立完成的）。但我认为有这样一个dataframe将是最有用的：

require(tidyverse)

data %>% 
  pivot_wider(names_from = Reason, names_prefix = "Reason_", values_from = Reason) %>% 
  mutate(across(.cols = -c(school_id, subject, vacancies), .fns = \(x) !is.na(x)))

# output
# A tibble: 4 × 6
  school_id subject vacancies Reason_Pay Reason_Location Reason_Experience
  <chr>     <chr>       <dbl> <lgl>      <lgl>           <lgl>            
1 1         English        40 TRUE       TRUE            FALSE            
2 2         Maths          20 TRUE       FALSE           FALSE            
3 3         French         40 FALSE      FALSE           TRUE             
4 3         French         10 FALSE      TRUE            FALSE

library(tidyr)

data |>
    pivot_wider(names_from = Reason,
                values_from = Reason,
                names_prefix = "Reason_") |>
    rename_with(.cols = starts_with("Reason_"),
                .fn =\(x) glue::glue("Reason_{1:3}"))

# A tibble: 4 × 6
  school_id subject vacancies Reason_1 Reason_2 Reason_3  
  <chr>     <chr>       <dbl> <chr>    <chr>    <chr>     
1 1         English        40 Pay      Location NA        
2 2         Maths          20 Pay      NA       NA        
3 3         French         40 NA       NA       Experience
4 3         French         10 NA       Location NA