R, 122 121 bytes

-1 byte thanks to Giuseppe

x=el(strsplit(scan(,""),""));n=seq(x);y=x>"/";for(i in n)cat(rep(" ",diffinv(y[n[-1]-1]+y[n[-1]]-1)[i]),x[i],"\n",sep="")

Try it online! (Note that el(x) does not work on TIO so I have replaced it with x[[1]] for one extra byte.)

With extra whitespace:

x = el(strsplit(scan(,""),""))
n = seq(x)
y = x>"/"
for(i in n) {
  cat(rep(" ", diffinv(y[n[-1]-1]+y[n[-1]]-1)[i]), x[i], "\n", sep="")
}

Explanation: This answer is based on the observation that the number of leading spaces changes each line by -1, plus the number of / in the previous and current lines.

If we have N slashes, the variable y is a vector of length N, with 1 for each position with \, 0 otherwise. Therefore, to get the change in number of leading spaces per line, we calculate y[1:(N-1)] + y[2:N] - 1. The function diffinv converts these differences into a sequence, starting with 0. The rest is just a matter of assembling each line as the required number of trailing spaces, followed by the relevant slash and a newline.

R, 122 121 bytes

-1 byte thanks to Giuseppe

x=el(strsplit(scan(,""),""));n=seq(x);y=x>"/";for(i in n)cat(rep(" ",diffinv(y[n[-1]-1]+y[n[-1]]-1)[i]),x[i],"\n",sep="")

Try it online!

With extra whitespace:

x = el(strsplit(scan(,""),""))
n = seq(x)
y = x>"/"
for(i in n) {
  cat(rep(" ", diffinv(y[n[-1]-1]+y[n[-1]]-1)[i]), x[i], "\n", sep="")
}

Explanation: This answer is based on the observation that the number of leading spaces changes each line by -1, plus the number of / in the previous and current lines.

If we have N slashes, the variable y is a vector of length N, with 1 for each position with \, 0 otherwise. Therefore, to get the change in number of leading spaces per line, we calculate y[1:(N-1)] + y[2:N] - 1. The function diffinv converts these differences into a sequence, starting with 0. The rest is just a matter of assembling each line as the required number of trailing spaces, followed by the relevant slash and a newline.

R, 122 bytes

x=el(strsplit(scan(,""),""));n=seq(x);y=x!="/";for(i in n)cat(rep(" ",diffinv(y[n[-1]-1]+y[n[-1]]-1)[i]),x[i],"\n",sep="")

Try it online! (Note that el(x) does not work on TIO so I have replaced it with x[[1]] for one extra byte.)

With extra whitespace:

x = el(strsplit(scan(,""),""))
n = seq(x)
y = x!="/"
for(i in n) {
  cat(rep(" ", diffinv(y[n[-1]-1]+y[n[-1]]-1)[i]), x[i], "\n", sep="")
}

Explanation: This answer is based on the observation that the number of leading spaces changes each line by -1, plus the number of / in the previous and current lines.

If we have N slashes, the variable y is a vector of length N, with 1 for each position with /, 0 otherwise. Therefore, to get the change in number of leading spaces per line, we calculate y[1:(N-1)] + y[2:N] - 1. The function diffinv converts these differences into a sequence, starting with 0. The rest is just a matter of assembling each line as the required number of trailing spaces, followed by the relevant slash and a newline.

R, 122 121 bytes

-1 byte thanks to Giuseppe

x=el(strsplit(scan(,""),""));n=seq(x);y=x>"/";for(i in n)cat(rep(" ",diffinv(y[n[-1]-1]+y[n[-1]]-1)[i]),x[i],"\n",sep="")

Try it online! (Note that el(x) does not work on TIO so I have replaced it with x[[1]] for one extra byte.)

With extra whitespace:

x = el(strsplit(scan(,""),""))
n = seq(x)
y = x>"/"
for(i in n) {
  cat(rep(" ", diffinv(y[n[-1]-1]+y[n[-1]]-1)[i]), x[i], "\n", sep="")
}

Explanation: This answer is based on the observation that the number of leading spaces changes each line by -1, plus the number of / in the previous and current lines.

If we have N slashes, the variable y is a vector of length N, with 1 for each position with \, 0 otherwise. Therefore, to get the change in number of leading spaces per line, we calculate y[1:(N-1)] + y[2:N] - 1. The function diffinv converts these differences into a sequence, starting with 0. The rest is just a matter of assembling each line as the required number of trailing spaces, followed by the relevant slash and a newline.