Jelly, 20 bytes

ṣ€”/ZṖF”#eƊ¿œiÐḟ”+ZE

A monadic Link accepting a list of lists of characters, [topic, pattern], which returns 1 or 0 for match or no-match respectively.

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How?

ṣ€”/ZṖF”#eƊ¿œiÐḟ”+ZE - Link: list of lists of characters, [topic, pattern]
 € - for each:
ṣ - split at occurrences of:
 ”/ - '/' character
 Z - transpose (any excess of topic is kept)
 ¿ - while...
 Ɗ - ...condition: last three links as a monad:
 ”# - '#' character
 e - exists in:
 F - flatten
 Ṗ - ...do: pop the tail off
 Ðḟ - filter discard those for which:
 œi - first multi-dimensional index of: ([] if not found, which is falsey)
 ”+ - '+' character
 Z - transpose
 E - all equal?

Ruby, 65 bytes

Regex solution. I added Regex.escape in case a criteria name just so happens to be something like com.java/string[]/n or something silly that would have regex pieces.

->s,cs=~/^#Regexp.escape(c).sub('#','.*').gsub'+','[^/]*'$/

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Non-regex solution, 77 bytes

Uses a nice simple split, zip, and match technique. I developed this one first before realizing that even with Regex.escape the regex solution would've been shorter anyways.

->s,cs.split(?/).zip(c.split ?/).all?i==j

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Perl 5 -pl, 50 bytes

$_="Q$_";s|\+|[^/]+|g;s/\#/.*/;$_=<>=~m|^$_$|

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Python 3, 72 bytes

lambda a,b:bool(re.match(b.translate(43:"[^/]+",35:".+"),a))
import re

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This problem can be trivially simplified to a regex match, though another more interesting method may produce better results.

EDIT I came up with a 107-byte solution not using regex. I don't know if it can get shorter than 72 or maybe I'm just not seeing to correct approach to this. Just the split-zip structure seems to be too large though. Try It Online!

Python 2, 85 84 80 92 89 bytes

lambda s,c:all(x in('+','#',y)for x,y in zip(c.split('/')+[0]*-c.find('#'),s.split('/')))

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Thanks to Jonathan Allan and Value Ink for pointing out bugs.

Haskell, 76 73 71 67 bytes

(a:b)#(c:d)=a=='+'&&b#snd(span(/='/')d)||a=='#'||a==c&&b#d
a#b=a==b

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Edit: -4 bytes thanks to @cole.

Clojure, 107 91 76 65 102 bytes

An anonymous function, returns subject topic as truthy and nil as falsey (valid in Clojure).

(defn ?[t c](every? #(#"#""+"(% 0)(% 1))(apply #(map vector % %2)(map #(re-seq #"[^/]+" %) [t c]))))

107 102 working
91 76 65 all defeated with regex chars

Kotlin, 106 bytes

fun f(s:List<String>)=s[1].split("/").filterIndexedi,v->v!="+"&&v!="#"&&v!=s[0].split("/")[i].count()==0

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Python 3, 99 88 bytes

Without using a regex.
With some help from Jonathan Allan and Chas Brown.

f=lambda s,p:p in(s,'#')or p[:1]in(s[:1],'+')and f(s[1:],p['+'!=p[:1]or(s[:1]in'/')*2:])

Charcoal, 36 bytes

≔⪪Ｓ/θ≔⪪Ｓ/ηＦ∧№η#⊟η≔…θＬηθＦ⌕Ａη+§≔θι+⁼θη

Try it online! Link is to verbose version of code. Outputs - (Charcoal's implicit output for true) for a match, nothing for no match. Explanation:

≔⪪Ｓ/θ

Split the subject on /s.

≔⪪Ｓ/η

Split the criteria on /s.

Ｆ∧№η#⊟η≔…θＬηθ

If the criteria contains (i.e. ends with) a # then remove it and trim the subject to the new length of the criteria.

Ｆ⌕Ａη+§≔θι+

Where the criteria contains + then replace that element in the subject with +.

⁼θη

Compare the subject with the criteria and implicitly print the result.

Retina 0.8.2, 42 bytes

%`$
/
+`^([^/]+/)(.*¶)(1|+/)
$2
^¶$|¶#/$

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%`$
/

Suffix a / to both lines.

+`^([^/]+/)(.*¶)(1|+/)
$2

Repeatedly remove the first element of both subject and criteria while they equal or the criteria element is a (happy) +.

^¶$|¶#/$

The criteria matches if it's just a # (with the / that was added earlier) otherwise both subject and criteria should be empty by this point.

Pyth, 22 bytes

:Q::E"[+]""[^/]+"#".*

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Jelly, 22 19 bytes

ḟ”+ṣ”/)ZẠƇṖœi”#$¿ZE

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A monadic link that takes as its argument [topic], [criterion] and returns 1 for a match and 0 for no match.

JavaScript, 69 66 bytes

t=>s=>new RegExp(s.split`+`.join`[^/]+`.split`#`.join`.+`).test(t)

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Python 3, 149 148 bytes

def f(t,c):t,c=t.split('/'),c.split('/');return all([c[i]=='+'or c[i]==t[i]or c[i]=='#'for i in range(len(c))])and not(len(c)!=len(t)and c[-1]!='#')

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05AB1E, 21 bytes

ε'/¡}ζʒ'+å≠}˜'#¡н2ôøË

Input as a list in the order [criteria, topic].

Try it online or verify all test cases.

Explanation:

ε # Map both strings in the implicit input-list to:
 '/¡ '# Split the string on "/"
 # i.e. ["+/+/A/B/#","z/y/A/B/x/w/v/u"]
 # → [["+","+","A","B","#"],["z","y","A","B","x","w","v","u"]]
 }ζ # After the map: zip/transpose the two string-lists,
 # with space as (default) filler
 # → [["+","z"],["+","y"],["A","A"],["B","B"],["#","x"],[" ","w"],
 # [" ","v"],[" ","u"]]
 ʒ } # Filter each pair by:
 '+å≠ '# Only keep those which do NOT contain a "+"
 # → [["A","A"],["B","B"],["#","x"],[" ","w"],[" ","v"],[" ","u"]]
 ˜ # Flatten the filtered list
 # → ["A","A","B","B","#","x"," ","w"," ","v"," ","u"]
 '#¡ '# Split the list by "#"
 # → [["A","A","B","B"],["x"," ","w"," ","v"," ","u"]]
 н # Only keep the first part
 # → ["A","A","B","B"]
 2ô # Split this back into pairs of two
 # → [["A","A"],["B","B"]]
 ø # Zip/transpose them back
 # → [["A","B"],["A","B"]]
 Ë # And check if both inner lists are equal
 # → 1 (truthy)
 # (after which the result is output implicitly)