Let's suppose you have a coefficient and a signal input value. If the coefficient has
\$F_C\$ fraction bits and the input has \$F_I\$ fraction bits then their product will have \$F_C + F_I \$ fraction bits. When you used 000000001 to represent the integer 1 you had implicitly set \$F_I = 0\$ so the products had the same format as the coefficients. If you use fixed-point values that are \$\ge 1.0\$ then you will need bits to the left of the binary point to represent the integer part of the value. As with the fraction bits, the number of integer bits in the product will equal the sum of the numbers of integer bits in the multiplier and multiplicand.
When you add fixed-point values they must have the same number of fraction bits (i.e. the binary point is aligned) and the sum will have the same number of fraction bits as the addends. If you don't have information about the actual range of values for the sum then you need to assume that a carry can occur, so you need an additional bit to the left of the binary point to represent the integer part of the number. That is, you need one more integer bit in the sum than the maximum number of integer bits in either of the addends.
In general, for the computer a string of bits is just a string of bits - you need to tell the computer what is represented there using what method. This is an important concept to understand. This is why no one can say for sure what is represented by 0xFF without the proper context.
For negative numbers there are a few common systems, with 2's complement being the most popular one.
Sign & Magnitude You use the MSB to represent the sign of the number. 0 is for positive, 1 is for negative. 0b00000010 is decimal 2, 0b10000010 is -2.
1's Complement You represent a binary number by the binary number you need to add to it in order to get all 1's (which means flipping all the bits). 0b0000010 is decimal 2, its 1's complement is 0b1111101. Now, do the same thing with the MSB as in S&M (sorry) and you get 0b00000010 for decimal 2 and 0b11111101 for -2. One of the issues with this system is having two representations for 0 (all 1's and all 0's).
2's Complement You represent a binary number by the binary number you need to add to it in order to get all 0's. You convert to 1's complement and add 1 - simple! MSB is used as sign indicator. so 0b00000010 is decimal 2, and 0b11111110 is -2.
Offset Binary This is long gone (I think) but basically you do 2's complement and invert the MSB used for the sign representation.
There are good reasons why 2's complement is the most commonly used system but I think that is outside the scope of this question.
Best Answer
Break down \$Y\$ such that \$Y=-y_3\cdot 2^3+\sum_{i=0}^2 y_i\cdot 2^i\$. But don't break down \$X\$. Just take it as a 2's complement signed value. Then \$X\cdot Y=y_0\cdot \underline{X\cdot 2^0}+y_1\cdot \underline{X\cdot 2^1}+y_2\cdot \underline{X\cdot 2^2}-y_3\cdot \underline{X\cdot 2^3}\$. Note that the first three lines are just a shifted version of \$X\$ (times an unsigned bit in \$Y\$) and, since \$X\$ is signed, the result must be sign-extended. The last term is again just a sign-extended result but in this case must be subtracted. This explains all four of the lines in case 1.
See above.
$$\begin{align*}X\cdot Y&=\left(-x_3\cdot 2^3+\sum_{i=0}^2 x_i\cdot 2^i\right)\cdot \left(-y_3\cdot 2^3+\sum_{i=0}^2 y_i\cdot 2^i\right) \\\\ &=x_3y_3\cdot2^6\\&\quad +\left(\sum_{i=0}^2 x_i\cdot 2^i\right)\cdot\left(\sum_{i=0}^2 y_i\cdot 2^i\right)\\&\quad-y_3\cdot\left(\sum_{i=0}^2 x_i\cdot 2^i\right)\cdot 2^3\\&\quad-x_3\cdot\left(\sum_{i=0}^2 y_i\cdot 2^i\right)\cdot 2^3 \\\\ &=x_3y_32^6+\sum_{i=0}^2 \sum_{j=0}^2 x_iy_j 2^{i+j}-y_32^3\sum_{i=0}^2 x_i 2^i-x_32^3\sum_{i=0}^2 y_i2^i \end{align*}$$
Or, in table form:
$$\begin{matrix} &2^7&2^6&2^5&2^4&2^3&2^2&2^1&2^0\\ \hline \\ +&0&x_3y_3&0&0&0&x_2y_0&x_1y_0&x_0y_0 \\\\ +&0&0&0&0&x_2y_1&x_1y_1&x_0y_1&0 \\\\ +&0&0&0&x_2y_2&x_1y_2&x_0y_2&0&0 \\\\ -&0&0&x_2y_3&x_1y_3&x_0y_3&0&0&0 \\\\ -&0&0&x_3y_2&x_3y_1&x_3y_0&0&0&0 \end{matrix}$$
Complement and increment the last two lines:
$$\begin{matrix} &2^7&2^6&2^5&2^4&2^3&2^2&2^1&2^0\\ \hline \\ +&0&x_3y_3&0&0&0&x_2y_0&x_1y_0&x_0y_0 \\\\ +&0&0&0&0&x_2y_1&x_1y_1&x_0y_1&0 \\\\ +&0&0&0&x_2y_2&x_1y_2&x_0y_2&0&0 \\\\ +&1&1&\overline{x_2y_3}&\overline{x_1y_3}&\overline{x_0y_3}&1&1&1 \\\\ +&0&0&0&0&0&0&0&1 \\\\ +&1&1&\overline{x_3y_2}&\overline{x_3y_1}&\overline{x_3y_0}&1&1&1 \\\\ +&0&0&0&0&0&0&0&1 \end{matrix}$$
Combine the constants:
$$\begin{matrix} &2^7&2^6&2^5&2^4&2^3&2^2&2^1&2^0\\ \hline \\ +&0&x_3y_3&0&0&0&x_2y_0&x_1y_0&x_0y_0 \\\\ +&0&0&0&0&x_2y_1&x_1y_1&x_0y_1&0 \\\\ +&0&0&0&x_2y_2&x_1y_2&x_0y_2&0&0 \\\\ +&0&0&\overline{x_2y_3}&\overline{x_1y_3}&\overline{x_0y_3}&0&0&0 \\\\ +&0&0&\overline{x_3y_2}&\overline{x_3y_1}&\overline{x_3y_0}&0&0&0 \\\\ +&1&0&0&1&0&0&0&0 \end{matrix}$$
Clear, now?