There are a plethora of (poorly-tagged) questions on math.SE about twos-complement binary numbers. In summary, the usual definition of twos complement is, to convert a negative number $-N$ to twos complement, you write $N$ (the magnitude of the number) in binary, "flip" (take the complement of) all the bits, then add $1$ (a number with $1$ in the least significant place and $0$ everywhere else).
The conversion in the reverse direction is: "flip" all the bits
and add one, treat the result as a positive number, convert it to decimal,
and then flip the sign of the number so that you get back the original
negative number. So, for example, $1011\,1101$
in eight-bit twos-complement is $-0100\,0010_2 +1 = -0100\,0011_2 = -67_{10}.$
For ones complement you just flip all the bits. So, for example, $1011\,1101$
in eight-bit ones-complement is $-0100\,0010_2 = -66_{10}.$
Personally, I prefer to think of a twos-complement negative number $-N$ in $w$ bits as simply $2^w - N$. The ones-complement representation of $-N$ is then
$(2^w - 1) - N$.
This is usually used in computer programming. There we usually store numbers in words of some size. As we write the numbers, the leftmost bit is the most significant. In unsigned binary with $8$ bit words we can store $0_{10}=00000000_2$ to $255_{10}=11111111_2$. The leftmost bit has a value of $128_{10}$ while all the others have less value, ranging from $64$ down to $1$.
When we borrow that for math, we do not use a fixed word length, so the most significant bit is the first $1$. We have $77_{10}=1001101_2$ and the most significant bit is the first $1$, with a value of $64$.
Best Answer
Everything is binary, leftshift of $10101$ denoted as $l(10101) = 101010$, note that $10*x = l(x)$. The basic idea is that you split one of the numbers up and use the distributive law. So your example would look like this:
\begin{eqnarray*} 1101 \cdot 110001 &=& 1101 \cdot (1 + 10000 + 100000) \\ &=& 1101 \cdot (1^0 + 10^4 + 10^5) \\ &=& 1101 \cdot 1^0 + 1101 \cdot 10^4 + 1101 \cdot 10^5 \\ &=& l^0(1101) + l^4(1101) + l^5(1101) \\ &=& 1101 + l(l(l(l(1101)))) + l(l(l(l(l(1101))))) \end{eqnarray*}