Why does this SHA-1 relation hold?
First let's write the original relation that they are deriving this one from:

$$
W_i = A_{i+1} - (A_i \lll 5) - f(A_{i-1}, A_{i-2} \ggg 2, A_{i-3} \ggg 2) - (A_{i-4} \ggg 2) - K_i
$$

So, if everything in the state bits are fixed except the ones in position $j+2$, all the operations on the fixed bits could be gathered together and we can simplify the relation to the variable ones. Now, let's derive the new relation.

For now, let's put the $c_i^j$ aside and write the term related to $f$. Then the variables in $A_{i-2}$ and $A_{i-3}$ (which are $a_{i-2}^{j+2}$ and $a_{i-3}^{j+2}$) are mixing with the fixed value of $A_{i-1}$ (which is $a_{i-1}^j$), and the rest is fixed. Thus the term $f(a_{i-1}^j, a_{i-2}^{j+2}, a_{i-3}^{j+2})$, and the result is in position $j$, so the $\ll j$ is added. We can add the term $(a_{i-4}^{j+2} \ll j)$ (corresponding to the last term in original relation).

The remaining terms can be collected in the $C_i^{j..0}$. Although both of the $A_{i+1}$ and $A_i$ have variable bits at position $j+2$, they do not contribute to the value of bit positions $j..0$ of $W$. The only way they get involved is that by left rotation of ($A_i \lll 5$), the variable bit gets rotated and falls in $j..0$. It happens if $(j+2) + 5 &gt; 31$, but they restrict the $j$ to $0\le j \lt 25$, so it doesn't happen.

And about the $E_i$ relations. Your relation is not quite right.
We know that:
$$
W_{i+16} = (W_{i+13} \oplus W_{i+8} \oplus W_{i+2} \oplus W_i) \lll s
$$
Therefore:
$$
W_{i+16} \ggg s = W_{i+13} \oplus W_{i+8} \oplus W_{i+2} \oplus W_i
$$
and in other words:
$$
W_{i+13} \oplus W_{i+8} \oplus W_{i+2} \oplus W_i \oplus (W_{i+16} \ggg s) = 0
$$
if there is no error in calculating $W_i$s.
So they define the error bit vectors as:
$$
E_i = W_{i+13} \oplus W_{i+8} \oplus W_{i+2} \oplus W_i \oplus (W_{i+16} \ggg s)
$$
with the goal of making all $E_i$ equal to zero.