How to pass the output of one command as the command-line argument to another?

It seems you could use a combination of the answers here. I'm guessing you are wanting to replace space chars with their escaped ascii values in the url. To do this, you need to replace them with "%20", not just "%". Here's a solution that should give you a complete answer:

$ wget `echo http://maps.google.be/maps\?saddr\=$1\&daddr\=$2 | sed -e 's/\ /\%20/g'` -q -O temp.html

The backticks indicate that the enclosed command should be interpreted first, and the result sent to wget. Notice I escaped the space and % chars in the sed command to prevent them from being misinterpreted. The -q option for wget prevents processing output from the command being printed to the screen (handy for scripting when you don't care about the in-work status) and the -O option specifies the output file. FYI, if you don't want to save the output to a file, but just view it in the terminal, use "-" instead of a filename to indicate stdout.

wget also accepts stdin with the - switch.

If you want to save the output in a file, use the -O switch.

echo http://maps.google.be/maps?saddr\=$1\&daddr\=$2 | sed 's/ /%/g' | wget -i- -O temp.html

You could use "xargs". A trivial example:

ls -1 *.c | sort -n | xargs cat

You would have to take care that xargs doesn't split its stdin into two or more invocations of the comman ("cat" in the example above).

you're not actually executing your url line :

#!/bin/sh
url="$(echo http://maps.google.be/maps?saddr\=$1\&daddr\=$2 | sed 's/ /%/g')"
wget $url