Let $M$ be a compact oriented $m$ manifold, and $F: M \to M$ a smooth map with fixed points. Then we can consider $\Delta = \{(x,x) \in M \times M\}$ and $G_{F}$ the graph of $F$ in $M \times M$. Both are embedded $m$ submanifold of $M \times M$. They have natural orientations induced by the obvious diffeomorphisms $i: M \to \Delta$ and $F: M \to M \times M$. The claim is that $\Delta \cap G_F$ has a natural orientation and it can be computed by sign of $det(I -dF)$.
How to derive this formula?
I should add that $\Delta \cap G_F$ is a zero manifold for dimension reasons. Then the orientation at each point is just an assignment of signs, so the formula does make sense.
I hope this will help. Choose an oriented basis $e_i$ for $M$. Under the map $M \to \Delta$ this basis will be sent to $(e_i,e_i)$ at the tangent space of $M\times M$. Under the map $F: M \to M \times M$ it will be sent to $(e_i, dF(e_i))$. As the intersection is transversal at point $p$, $\{(e_i,e_i)\}\cup\{(e_i,dF(e_i))\}$ is a basis for $T_{(p,p)}M\times M$. To calculate it’s orientation relative to the standard one $\{(e_i,0)\}\cup\{(0,e_i)\}$ is calculating: $$\left|\begin{pmatrix}I & I \\ dF & I \end{pmatrix}\right|=\det(I-dF)$$