Understanding the summing-point constraint
suppose that \$V_{DIFF} = 0.1\,v\$ initially, and then \$V_{OUT}=-10\,000\,v\$. If a fraction of this is sent back, say \$0.1\%\$.


The "fraction" that is sent back is not really a voltage, it's a current. 

If \$V_{OUT}\$ is -10,000 V, then there must flow through \$R_F\$ a current of \$\frac{10^4}{R_F}\mathrm{A}\$. Since the input impedance of the op-amp is very high (infinite in the ideal case), the resulting feedback voltage at the inverting input is determined by a voltage divider between \$R_F\$ and \$R_{in}\$. So the feedback voltage is \$V_{OUT}\frac{R_{in}}{R_F+R_{in}}\$. It's likely to be much higher than the 0.1% you assumed in your example.

How to solve for the output voltage including the forward gain and feedback terms, is explained in a previous question.