Understanding Convolution
I think the reason you are getting confused is that you are only considering the convolution of two signals, whereas convolution is not usually between signals but between a signal and an impulse response.

The best way to visualise it is probably with a filter.  For example, in 


we can see that the input comes in, and is multiplied by $b_0$ to give its contribution to the output $y$.  Then in the next time step, that input value is multiplied by $b_1$ to give its contribution to the output, and so forth.  In this case your $k$ variable specifies which time in the output sequence you want.  The output signal is given by $y[n] = (x * b)[n]$.

For example, consider the case of $n=4$ (that's the same as $k=4$ in your example).  That means we're looking at the 4th time step (since time = 0).  So the value at the input is $x[4]$, the fourth input value; the value to the right of the first $z^{-1}$ block is the previous input value $x[3]$; the value to the right of the next $z^{-1}$ block is $x[2]$; and so forth.  The output $y[4]$ will be given by 
$$y[4] = x[4]b_0 + x[3]b_1 + x[2]b_2 + x[1]b_3 + \cdots + x[-3]b_7 + \cdots$$
 which you should be able to see is the discrete convolution sum, and includes the product of $x[-3]$ with $b[7]$.

Any linear filter or transfer function can be implemented as a convolution, where the output is given by the convolution of the input with the system impulse response.  Try reading up on convolution in a text book, and look particularly at the sliding-strip method of calculating it, which might help with your visualisation.