JavaScript (Node.js), 61 55 54 bytes

Saved 1 byte thanks to @nwellnhof

Takes input as an array of characters.

s=>s.map(p=c=>r+=p>(p=(+c?~c:1-Buffer(c)[0])%6),r=0)|r

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How?

For all characters but digits greater than $0$, the 0-indexed column $x$ is given by:

$$x=(c-1)bmod 6$$

where $c$ is the ASCII code of the character.

For positive digits $n$, we need to do instead:

$$x=(n+1)bmod 6$$

Examples:

"a" --> (97 - 1) mod 6 = 96 mod 6 = 0

"b" --> (98 - 1) mod 6 = 97 mod 6 = 1

"0" --> (48 - 1) mod 6 = 47 mod 6 = 5

"3" --> ( 3 + 1) mod 6 =  4 mod 6 = 4

Commented

s =>                       // s = input string (as array)

  s.map(p =                // initialize p to a non-numeric value

  c =>                     // for each character c in s:

    r +=                   //   update the result r:

      p > (                //   compare p with

        p = (              //   the new value of p defined as:

          +c ?             //     if c is a positive digit:

            ~c             //       -(int(c) + 1)

          :                //     else:

            1-Buffer(c)[0] //       -(ord(c) - 1)

        ) % 6              //     apply modulo 6

      ),                   //   yields 1 if the previous value is greater than the new one

    r = 0                  //   start with r = 0

  ) | r                    // end of map(); return r

Jelly, 11 bytes

⁾04yO‘%6<ƝS

A monadic Link accepting a list of (uppercase) characters.

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How?

First replaces any '0's with '4's (so the rest of the code treats them as being in the rightmost column). Then casts to ordinals, adds one and modulo's by 6 to get 0-based column indices. Then compares neighbours with is-less-than and sums the result.

⁾04yO‘%6<ƝS - Link: list of characters         e.g. "BLAST20"

⁾04         - list of characters = ['0', '4']

   y        - translate                             "BLAST24"

    O       - ordinals                              [66,76,65,83,84,50,52]

     ‘      - increment                             [67,77,66,84,85,51,53]

       6    - literal six

      %     - modulo                                [ 1, 5, 0, 0, 1, 3, 5]

         Ɲ  - neighbourly:

        <   -   less than?                          [  1, 0, 0, 1, 1, 1  ]

          S - sum                                   4

Perl 6, 45 39 bytes

{sum .[1..*]Z<$_}o{(2 X-.ords)X%46 X%6}

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Works with uppercase letters. (2-ord(c))%46%6 computes the reversed x coordinate.

K (ngn/k), 32 31 bytes

+/>':(+6 6#,/"a10"+!'26 9 1)?0+

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Clean, 85 bytes

import StdEnv

(s=sum[1\a<-s&b<-tl s|b>a])o map(e=(toInt e+if(e-'1'>'/')1 -1)rem 6)

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Ruby, 56 bytes

->s{w=9;s.count{|c|w<w=[*?a..?z,*?1..?9,?0].index(c)%6}}

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Preliminary naive version, will be golfed.

Japt -x, 14 bytes

®rT4 c Ä u6Ãä<

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Port of this Jelly answer. Takes input as an array of characters, with letters uppercase.

Explanation:

®rT4 c Ä u6Ãä<    :

®          Ã      :Map each character through:

 rT4              : Replace 0 with 4

     c            : Get the char-code

       Ä          : Increment it

         u6       : Modulo 6

            ä<    :Replace with 1 if you had to move right, 0 otherwise

                  :Implicitly sum and output

Java (OpenJDK 8), 73 bytes

Not a bad solution for Java! That zero being on the right-hand side cost me several bytes.

t->{int a=9,c=0;for(int d:t)c+=a<(a=(--d+(d/48==1?2:0))%6)?1:0;return c;}

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Explained

t -> {                          // Lambda taking a char array as input

    int a=9,                    // Initialise last column value

        c=0;                    // Initialise frustration count

    for(int d:t)                // Loop through all chars in title

        c+=                     // increment the frustration count if...

          a<                    // The last column is smaller than the current column

            (a=                 // Set last column to current column

              (--d+             // Decrement ascii value of char

                  (d/48==1      // If ascii decremented ascii value is between 48 and 95

                    ?2:0)       // increment by 2 (1 total) or 0 (-1 total)

                )%6)            // Mod 6 to retrieve column index

            ?1:0;               // Increment if to right hand side

    return c;                   // return calculated frustration count

}

05AB1E, 12 11 bytes

-1 byte thanks to @Kevin Cruijssen

¾4:Ç>6%¥1@O

Another port of Jonathan Allan's Jelly answer. Takes input in uppercase.

Explanation:

¾4:Ç>6%¥1@O   //full program

¾4:           //replace all '0's with '4's

   Ç          //get ASCII code points

    >         //increment

     6%       //modulo 6

       ¥      //get deltas

        1@    //is >= 1

          O   //sum

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Retina 0.8.2, 46 bytes

T`l1-90`1-61-61-61-61-61-6

.

;$&$*

&`;(1+);11

Try it online! Link includes test cases. Explanation:

T`l1-90`1-61-61-61-61-61-6

List the alphabet and digits in the order on the OSK and map each one to a (1-indexed) column number.

.

;$&$*

Convert each column number to unary.

&`;(1+);11

Count the number of columns that are followed by a larger (i.e. rightwards) column. The &` allows the matches to overlap.

Python 2, 84 bytes

def f(s,k=6):j=(ord(s[0])+('0'<s[0]<':')*2-1)%6;return(j>k)+(~-len(s)and f(s[1:],j))

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Mathematica, 102 bytes

Differences[Last@@Join[Alphabet,ToString/@Range@9,{"0"}]~Partition~6~Position~#&/@#]~Count~_?(#>0&)&

Pure function. Takes a list of characters as input and returns a number as output. This is a pretty naive solution, golfing suggestions welcome.

C (gcc),  82  79 bytes

o;c(i){i+=i<60&&i>48?1:5;i%=6;}f(char*s){for(o=0;*++s;o+=c(*s)>c(*(s-1)));o=o;}

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This function will only support lowercase inputs

Ungolfed and commented:

o; //Used for output

c(i){             //Calculates the column of given character

     i+=          //Correct i to get the correct column

        i<60      //If i is a digit...

        && i>48   //... but not '0'

        ?1           //Then move it one column on the right

        :5;          //Else move it five columns on the right

     i%=6;        //Get the column number

}

f(char*s){                        // The actual "frustrating" function

          for(                    //Loop for each character

              o=0;                //reinitialize output

              *++s;               //move to next character / while this is not ''

              o+=c(*s)>c(*(s-1))  //Increment if current character is on the right of the previous one

             );

           o=o;                   // Outputs result

}

If my function is allowed to accept wide character strings, it can be reduced to 78 bytes with:

o;c(i){i+=i<60&&i>48?1:5;i%=6;}f(int*s){for(o=0;*++s;o+=c(*s)>c(*(s-1)));o=o;}

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This version just accept input as int* instead of char*

Edits:

• Golfed 3 bytes in the calculation of the column (function c)