R, 138 bytes

I'll bet there's an easier way to do this. Use gsub to get the alternating numeric positions.

function(x,r=-48+~x)Reduce(paste0,ifelse(58<~gsub("[123]{2}","0a",x),"123"["一二三",r],'0-9'["〇〡-〩",r]))

"~"=utf8ToInt

"["=chartr

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JavaScript, 81 bytes

s=>s.replace(/./g,c=>(p=14>>c&!p)|c>3?eval(`"\u302${c}"`):'〇一二三'[c],p=0)

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Using 14>>c saves 3 bytes. Thanks to Arnauld.

Retina, 46 bytes

/[1-3]{2}|./_T`d`〇〡-〩`^.

T`123`一二三

Try it online! Link includes test cases. Explanation:

/[1-3]{2}|./

Match either two digits 1-3 or any other digit.

_T`d`〇〡-〩`^.

Replace the first character of each match with its Suzhou.

T`123`一二三

Replace any remaining digits with horizontal Suzhou.

51 bytes in Retina 0.8.2:

M!`[1-3]{2}|.

mT`d`〇〡-〩`^.

T`¶123`_一二三

Try it online! Link includes test cases. Explanation:

M!`[1-3]{2}|.

Split the input into individual digits or pairs of digits if they are both 1-3.

mT`d`〇〡-〩`^.

Replace the first character of each line with its Suzhou.

T`¶123`_一二三

Join the lines back together and replace any remaining digits with horizontal Suzhou.

Perl 5 -pl -Mutf8, 53 46 bytes

-7 bytes thanks to Grimy

s/[123]{2}|./OS&$&/ge;y//〇〡-〰一二三/c

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Explanation

# Binary AND two consecutive digits 1-3 (ASCII 0x31-0x33)

# or any other single digit (ASCII 0x30-0x39) with string "OS"

# (ASCII 0x4F 0x53). This converts the first digit to 0x00-0x09

# and the second digit, if present, to 0x11-0x13.

s/[123]{2}|./OS&$&/ge;

# Translate empty complemented searchlist (0x00-0x13) to

# respective Unicode characters.

y//〇〡-〰一二三/c

Java (JDK), 120 bytes

s->{for(int i=0,p=0,c;i<s.length;)s[i]+=(p>0&p<4&(c=s[i++]-48)>0&c<4)?"A䷏乚䷖".charAt(c+(p=0)):(p=c)<1?12247:12272;}

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Credits

• -3 bytes thanks to Kevin Cruijssen
  

JavaScript (ES6),  95 89  88 bytes

Saved 6 bytes thanks to @ShieruAsakoto

Takes input as a string.

s=>s.replace(i=/./g,c=>'三二一〇〡〢〣〤〥〦〧〨〩'[i=112>>i&c<4?3-c:+c+3])

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Python 3, 102 bytes

f=0

for i in input():f=i in'123'and 9-f;print(end='〇一二三〤〥〦〧〨〩〡〢〣'[int(i)+f])

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mypetlion reminded me of a trivial golf. -4 bytes.

Clean, 181 165 bytes

All octal escapes can be replaced by the equivalent single-byte characters (and are counted as one byte each), but used for readability and because otherwise it breaks TIO and SE with invalid UTF-8.

import StdEnv

u=mapc={'343','200',c}

?s=((!!)["〇":s++u['244245246247250']])o digitToInt

$=

$[h:t]=[?(u['241242243'])h:if(h-'1'<'03')f$t]

f=

f[h:t]=[?["一","二","三"]h: $t]

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An encoding-unaware compiler is both a blessing and a curse.

Perl 6 -p, 85 61 bytes

-13 bytes thanks to Jo King

s:g[(1|2|3)<((1|2|3)]=chr $/+57;tr/0..</〇〡..〩一二三/

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Red, 198 171 bytes

func[n][s: charset"〡〢〣"forall n[n/1: either n/1 >#"0"[to-char 12272 + n/1][#"〇"]]parse

n[any[[s change copy t s(pick"一二三"do(to-char t)- 12320)fail]| skip]]n]

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Jelly, 38 bytes

9Rż“øƓ“œ%“øƈ’;-25+⁽-EỌœị@DżD<4«Ɗ‘×¥ƊƊ

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C, 131 bytes

f(char*n){char*s="〇〡〢〣〤〥〦〧〨〩一二三",i=0,f=0,c,d;do{c=n[i++]-48;d=n[i]-48;printf("%.3s",s+c*3+f);f=c*d&&(c|d)<4&&!f?27:0;}while(n[i]);}

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Explanation:
First of all - I'm using char for all variables to make it short.

Array s holds all needed Suzhou characters.

The rest is pretty much iterating over the provided number, which is expressed as a string.

When writing to the terminal, I'm using the input number value (so the character - 48 in ASCII), multiplied by 3, because all these characters are 3 bytes long in UTF-8.
The 'string' being printed is always 3 bytes long - so one real character.

Variables c and d are just 'shortcuts' to current and next input character(number).

Variable f holds 0 or 27 - it says if the next 1/2/3 character should be shifted to alternative one - 27 is the offset between regular and alternative character in the array.

f=c*d&&(c|d)<4&&!f?27:0 - write 27 to f if c*d != 0 and if they are both < 4 and if f isn't 0, otherwise write 0.

Could be rewritten as:

if( c && d && c < 4 && d < 4 && f == 0)

f = 27

else

f = 0

Maybe there are some bytes to shave off, but I'm no longer able to find anything obvious.

Ruby -p, 71 bytes

$_=gsub(/[1-3]K[1-3]/){|x|(x.ord+9).chr}.tr"0-<","〇〡-〩一二三"

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K (ngn/k), 67 bytes

{,/(0N 3#"〇一二三〤〥〦〧〨〩〡〢〣")x+9*<x&x<4}@10

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10 get list of decimal digits

{ }@ apply the following function

x&x<4 boolean (0/1) list of where the argument is less than 4 and non-zero

< scan with less-than. this turns runs of consecutive 1s into alternating 1s and 0s

x+9* multiply by 9 and add x

juxtaposition is indexing, so use this as indices in...

0N 3#"〇一二三〤〥〦〧〨〩〡〢〣" the given string, split into a list of 3-byte strings. k is not unicode aware, so it sees only bytes

,/ concatenate

Jelly, 35 bytes

9Ḷ;-26ż“/Ẉ8‘+⁽ȷc¤ṃ@ɓD_2ỊŒgÄFị"+⁽-FỌ

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Wolfram Language (Mathematica), 117 bytes

FromCharacterCode[12320+(IntegerDigits@#/. 0->-25//.MapIndexed[{a___,c=#2[[1]],c,b___}->{a,c,#,b}&,{0,140,9}+7648])]&

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Note that on TIO this outputs the result in escaped form. In the normal Wolfram front end, it will look like this:

Japt, 55 bytes

s"〇〡〢〣〤〥〦〧〨〩"

ð"[〡〢〣]" óÈ¥YÉîë2,1Ãc

£VøY ?Xd"〡一〢二〣三":X

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It's worth noting that TIO gives a different byte count than my preferred interpreter, but I see no reason not to trust the one that gives me a lower score.

Explanation:

    Step 1:

s"〇〡〢〣〤〥〦〧〨〩"        Convert the input number to a string using these characters for digits



    Step 2:

ð                            Find all indexes which match this regex:

 "[〡〢〣]"                    A 1, 2, or 3 character

           ó    Ã            Split the list between:

            È¥YÉ              Non-consecutive numbers

                  ®    Ã     For each group of consecutive [1,2,3] characters:

                   ë2,1      Get every-other one starting with the second

                        c    Flatten



    Step 3:

£                              For each character from step 1:

 VøY                           Check if its index is in the list from step 2

     ?                         If it is:

      Xd"〡一〢二〣三"            Replace it with the horizontal version

                     :X        Otherwise leave it as-is

C# (.NET Core), 107 bytes, 81 chars

n=>{var t="〇一二三〤〥〦〧〨〩〡〢〣";var b=0;return n.Select(k=>t[k+(b+=k>0&k<4?1:b)%2*9]);}

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Saved 17 bytes thanks to @Jo King

Old Answer

C# (.NET Core), 124 bytes, 98 chars

n=>{var t="〇一二三〤〥〦〧〨〩〡〢〣";var b=0<1;return n.Select(k=>{b=k>0&k<4?!b:0<1;return b?t[k]:t[k+9];});}

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Takes input in the form of a List, and returns an IEnumerable. I don't know if this input/output is ok, so just let me know if it isn't.

Explanation

How this works is that it transforms all the integers to their respective Suzhou numeral form, but only if variable b is true. b is inverted whenever we meet an integer that is one, two, or three, and set to true otherwise. If b is false, we turn the integer to one of the vertical numerals.

R, 104 bytes

function(x,`[`=chartr)"a-jBCD"["〇〡-〩一二三",gsub("[bcd]\K([bcd])","\U\1","0-9"["a-j",x],,T)]

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An alternative approach in R. Makes use of some Perl-style Regex features (the last T param in substitution function stands for perl=TRUE).

First, we translate numerals to alphabetic characters a-j, then use Regex substitution to convert duplicate occurrences of bcd (formerly 123) to uppercase, and finally translate characters to Suzhou numerals with different handling of lowercase and uppercase letters.

Credit to J.Doe for the preparation of test cases, as these were taken from his answer.

C#, 153 bytes

n=>Regex.Replace(n+"",@"[4-90]|[1-3]{1,2}",x=>"〇〡〢〣〤〥〦〧〨〩"[x.Value[0]-'0']+""+(x.Value.Length>1?"一二三"[x.Value[1]-'0'-1]+"":""))

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