Jelly, 6 bytes

n⁶Sċ2Ẓ

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Haskell, 34 bytes

a#b=[1|(p,'-')<-zip a b,p>' ']>[1]

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I got this one by combining my below technique with the other haskell answer.

Haskell, 45 bytes

x#'-'|x/=' '=1
x#y=0
(((>1).sum).).zipWith(#)

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This counts the number of body parts (non-space characters) that are on top of the platform and then checks it is greater than 1. The reason we count body parts on the platform rather than body parts off is that zipWith will chop the top line to be the length of the bottom and thus can chop off Jimmy's body parts. This prevents us from having to do something like cycle" " to pad the list.

Python 2, 42 37 bytes

lambda j,p:'--'in p[j.find('/'):][:3]

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5 bytes thx to negative seven

Python 3, 88 43 bytes

Input is given in the form of a list containing two strings: the first string is the first line; the second string is the second line.

lambda a:sum(1-(" "in i)for i in zip(*a))>1

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Another version, tying for 43 bytes (I haven't been able to get it shorter than 43):

lambda a,b:b[a.find("/"):][:3].count("-")>1

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Down by 42 bytes thanks to a tip from Jo King.

Old Version:

lambda s:sum((s.split("n")[1]+" "*len(s))[i]=="-"and s[i]!=" "for i in range(len(s)))>1

-2 bytes thanks to Sriotchilism O'Zaic.

This works by taking two separate inputs, then pairing up corresponding letters. It counts the number of pairs where neither character is a space, then returns True if that number is greater than 1.

JavaScript (ES6), 38 bytes

Takes input as (a)(b). Returns $0$ or $1$.

a=>b=>b[a.search`o`]=='-'&/--/.test(b)

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How?

We look for the position of the middle part "o" of Jimmy's body in the first string and test wether there's a dash in the second string at the same position.

b[a.search`o`] == '-'

The only case where Jimmy would be unsafe in this situation is with a single-dash platform:

/o
 -

So we additionally make sure that the platform has a width of at least $2$:

/--/.test(b)

JavaScript (ES6), 36 bytes

Alternate version if we assume that there's always either dashes or spaces below Jimmy (i.e. the input is rectangular).

a=>b=>b[a.search`o`]!=0&/--/.test(b)

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Takes advantage of the fact that the coercion to a numeric value is $0$ for a space and NaN for a dash.

Haskell, 59 bytes

f a b=sum[1|(p,q)<-zip a$b++cycle" ",elem p"/o\",q==' ']<2

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The function is called like so: f "/o\ " " -- "

How it works (for f "/o\" " -"):

b++cycle" " - Adds an infinite number of spaces after b to ensure that Jimmy is always above a - or (" -" → " - ..."

zip a$b++cycle" " - Zips the two strings together ([('/',' '), ('o','-'), ('\',' ')])

(p,q)<-zip a$b++cycle - For each pair in the zipped list

[1|(p,q)<-zip a$b++cycle" ",elem p"/o\",q==' '] - Generates a list of 1s, whose length is the number of pairs satisfying the conditions:

elem p"/o\" - The character in the top string is one of Jimmy's body parts. (Satisfied by all three pairs in this example)

q==' ' - The character in the bottom string is a space. (Satisfied by ('/', ' ') and ('\', ' '))

So, the pair has to be one where one of Jimmy's body parts is above a space.

Because in this example, two pairs satisfy both conditions, the list is [1,1]

sum[1|(p,q)<-zip a$b++cycle" ",elem p"/o\",q==' '] - Take the sum of those 1s (i.e. the length of the list), which in this example is 2.

sum[1|(p,q)<-zip a$b++cycle" ",elem p"/o\",q==' ']<2 - Check if the number of body parts above a space is less than 2. In this example, it's not, so Jimmy will fall off. :(

Dyalog APL Extended, 11 10 bytes

2≤+/∧⌿<⍵

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Explanation:

2≤+/∧⌿<⍵ a monadic function with argument ⍵
 <⍵ Compare with the implicit prototype element - a space.
 Returns a matrix of characters that are greater than 0x20
 ∧⌿ and-reduce, i.e. places where both Jimmy and a platform is
 +/ sum that array, i.e. amount of body parts over a platform
 2≤ is that greater-or-equal to 2?

Perl 6, 18 bytes

Bq/o&[~^]

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Takes a two parameters and returns a boolean of whether Jimmy will stay on the platform. This works by XORing the two lines together and checking if either part of Jimmy is still on the platform.

Explanation:

 &[~^] # String XOR operator
 o # Combined with the anonymous function
 ?/ / # That checks for the regex match
 ''B # Unprintable, B, which is "/o" ~^ "--"
 |Bq # Or B, q, which is "o" ~^ "--"

VBA (Excel), 87 bytes EXCEL, 94 bytes

A1 = Jimmy , A2 = platform

=(FIND("-",A2)-FIND("/",A1)<2)*(FIND("",A1)-LEN(A2)<2)*(LEN(A2)-LEN(SUBSTITUTE(A2,"-",""))>1)

?[(FIND("-",A2)-FIND("/",A1)<2)*(FIND("",A1)-LEN(A2)<2)]*(len(replace([A2]," ",""))>1)

R, 35 bytes

function(x)sum(colSums(x!=" ")>1)>1

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Based on @EriktheOutgolfer’s excellent Jelly answer so please upvote that one too!

Input is a 2-d matrix of characters.

C (gcc), 103 bytes

i,j,l;f(char*s)for(i=0;*s-10;j=*s++-47?j:i)++i;l=strlen(s)>j+1;s+=j;return l&&*s+s[1]+(s[2]?:32)>109;

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Would be much shorter (75 bytes), if it could be assumed that there is trailing whitespace after the platform.

Retina, 31 bytes

^(.)*/.*¶(?<-1>.)*(?(1)!) ?---?

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A simple regex that determines whether at least two pieces of the platform are directly beneath Jimmy. Assumes that the input will contain no ! characters.

Japt -¡, 9 bytes

-1 byte if the input can be reversed.

VtUao)è-²

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V, 18 bytes

0vt/"_dj.d3lVkp0#x

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Output whitespace only if jimmy fall of the platform. Output something non-whitespace if jimmy stay on the platform.

Japt, 8 bytes

Õ·kèS Êz

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