Significant Whitespace

Python, 54 + 56 = 110 bytes

Counter:

m=lambda x:sum(y.isspace()for y in x)
+1
0<9or x.split()

Splitter:

m=lambda x:sum(y.isspace()for y in x)+10<9or x.split()

For the counter, we use the fact that Python is okay with having just an expression on a line. It's necessary to split up +1 and 0<9or x.split() to stop a NameError from being thrown, as 0<9 being True prevents x.split() from being evaluated due to short circuiting.

For the splitter, since the number of whitespace is always nonnegative, sum(y.isspace()for y in x)+10<9 is always False and the splitting function comes into play.

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Alternative 1, 59 + 60 = 119 bytes

Counter:

m=lambda x:[x.split(),sum(y.isspace()for y in x)][min([1])]

Splitter:

m=lambda x:[x.split(),sum(y.isspace()for y in x)][m in([1])]

The results of both counting and splitting are stored in a two-element list. The list is indexed by either min([1]), returning the minimum of the one-element list containing 1, or m in([1]), which returns False (equivalent to 0) as m is not contained in [1].

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Alternative 2, 67 + 69 = 136 bytes

Counter:

ted=1;s=lambda x:[x.split(),sum(y.isspace()for y in x)][not s or ted]

Splitter:

ted=1;s=lambda x:[x.split(),sum(y.isspace()for y in x)][not sorted]

Like above, the results of both counting and splitting are stored in a two-element list. sorted is a built-in function which is a truthy value, so not sorted returns False (equivalent to 0). For not s or ted, since s is a function and also truthy, not s is False and ted = 1 is returned.

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Alternative 3, 59 + 60 = 119 bytes

Counter:

def f(s):a=s.split();a1=len(s)-len((s*0).join(a));return a1

Splitter:

def f(s):a=s.split();a1=len(s)-len((s*0).join(a));return a
1

This is a function where splitter's result is stored in the variable a, and counter's result is stored in the variable a1. Like before, Python is fine with having just an expression on a line, in this case 1. Splitting up a1 determines what to return from the function.

Pyth, 16 + 15 = 31 bytes

Try it here.

Counter:

L@,cb)sm!cd)b1 0

Splitter:

L@,cb)sm!cd)b10

These each define a function, y, which takes a string input to solve the desired task.

Thanks to @FryAmTheEggman for the idea of using Pyth's feature of modular indexing into lists to shave a character.

Test cases:

L@,cb)sm!cd)b1 0y++"abc def"b"gh ij k"
L@,cb)sm!cd)b10y++"abc def"b"gh ij k"

Explanation:

L                  define a function, y, which takes one input, b.
 @                 Index into
  ,                2-tuple of
   cb)             b.split()                          (solution to splitter)
   s               sum over                           (solution to counter)
    m              map, with input d, to
     !cd)          logical negation of d.split()      (empty list is falsy)
     b             over b.
                   Index is either:
   10
   1
                   Indexing is modulo the length of the list in Pyth.
 0                 In one case, a 0 with a leading space is outside the function.
                   Leading space suppresses print, so the 0 is invisible.

Java 8, 239 + 240 = 479

Count the whitespace (returns Integer)

Object f(String s){String r=new String(new char[]{92,'s'}),e=new String();int intx=0;intx=1;return intx>0?s.chars().filter(c->c==9|c==10|c==32).count():java.util.Arrays.stream(s.split(r)).map(t->t.replaceAll(r,e)).filter(t->t.length()>0);}

Split on the whitespace (returns Stream<String>)

Object f(String s){String r=new String(new char[]{92,'s'}),e=new String();int intx=0;int x=1;return intx>0?s.chars().filter(c->c==9|c==10|c==32).count():java.util.Arrays.stream(s.split(r)).map(t->t.replaceAll(r,e)).filter(t->t.length()>0);}

Explanation:

Object f(String s){
    String r=new String(new char[]{92,'s'}),e=new String();  // init regex'es

    int intx=0;     // critical variable

    intx=1;         // change intx to 1
              OR
    int x=1;        // new, unused variable

    return intx>0 ? // test variable to decide what to do
      s.chars().filter(c->c==9|c==10|c==32).count() :
      java.util.Arrays.stream(s.split(r)).map(t->t.replaceAll(r,e)).filter(t->t.length()>0);
}