What does it mean geometrically that the diagonal cohomology class is concentrated along the diagonal? In Milnor Stasheff's Characteristic Classes, the geometric interpretation of Lemma 11.8 on P125 is that the diagonal cohomology class is "concentrated along the diagonal". The lemma states that

For any cohomology class $a \in H^*(M)$, the product $(a \times 1) \cup u''$ is equal to $(1 \times a) \cup u''$, in $H^*(M \times M)$.

Here, $u''$ is the "diagonal cohomology class", defined as in https://mathoverflow.net/a/74199/42662, or alternatively the image of the fundamental class of $H^n(M \times M, M \times M - \Delta(M)) \to H^n(M \times M)$.
How does the formal statement in the lemma give rise to the geometric "concentration" interpretation?
 A: It means that if $\alpha, \beta \in H^*(M \times M)$ are two forms which are equal when restricted to the diagonal, then their pairing with the diagonal class $\int_{M \times M} \alpha u'' = \int_{M \times M} \beta u''$ are equal. So in a sense only the value of $u''$ when restricted to the diagonal counts. (You also need Poincaré duality to be able to say that knowing all pairings of $u''$ with all other classes is enough to know $u''$.)
Indeed, by Künneth's formula, over a field you have $H^*(M \times M) = H^*(M) \otimes H^*(M)$. So you can decompose $\alpha = \sum_i \alpha'_i \otimes \alpha''_i$ and $\beta = \sum_j \beta'_j \otimes \beta''_j$. The restriction to the diagonal is just the cup product: if $\delta : M \to M \times M$ is the inclusion of the diagonal ($\delta(x) = (x,x)$) then $\delta^*\alpha = \sum_i \alpha'_i \alpha''_i$ and similarly for $\beta$. Suppose $\delta^*\alpha = \delta^*\beta$. Then you get:
$$\begin{align}
\int_{M \times M} \alpha u''
& = \sum_i \int_{M \times M} (\alpha'_i \otimes \alpha''_i) u'' \\
& = \int_{M \times M} (\sum_i \alpha'_i \alpha''_i \otimes 1) u'' \\
& = \int_{M \times M} (\sum_j \beta'_j \beta''_j \otimes 1) u'' \\
& = \int_{M \times M} \beta u''.
\end{align}$$
Here I used $(a \otimes b) u'' = (a \otimes 1) (1 \otimes b) u'') = (a \otimes 1) (b \otimes 1) u'' = (ab \otimes 1) u''$.
Or for something more "geometric", if you work with smooth manifolds and differential forms: if you choose any neighborhood of the diagonal, no matter how small, you will be able to find a representative of $u''$ which vanishes outside this neighborhood.
