Why doesn't echo called as /bin/sh -c echo foo output anything?

From man sh

-c string   If the -c option is present, then commands are read from string. 
            If there are arguments after the string, they are assigned to the
            positional parameters, starting with $0

It means your command should be like this:

 $ sh -c 'echo "$0"' foo 
 foo

Similarly:

$ sh -c 'echo "$0 $1"' foo bar
foo bar

That was the first part to understand; the second case is simple and doesn't need explanation, I guess.

$ echo foo
foo

This calls echo with the argument foo and foo is printed.

$ /bin/sh -c echo foo

This invokes the shell with the argument echo and provides foo as argument $0. The echo outputs a new line and you discard the foo. If you want to output foo, quote the argument:

sh -c 'echo foo'

or use the provided argument:

sh -c 'echo $0' foo

In this example

$ /bin/sh -c 'echo foo; echo bar'
foo
bar

The shell is invoked with the argument echo foo; echo bar which outputs

foo
bar

In this command:

echo foo

echo is the binary (or built-in command) and foo is the first argument.

Here:

/bin/sh -c echo foo

/bin/sh is the binary, whose first argument is -c, which itself accepts a "command string" as parameter. This is echo in the above example. Then there is a third argument: foo which is an argument for /bin/sh, not for echo. That's why in your third example:

/bin/sh -c 'echo foo; echo bar'

... both are printed. You quoted the argument. Thus: the first argument is -c, and the parameter to this argument is 'echo foo; echo bar' which is interpreted whole as one argument; as the "command string".