A simply connected $4$-manifold is spin iff all embedded oriented surfaces have even self-intersection number or, equivalently, if the quadratic form $H_2 (M;Z) \to Z$ induced by the intersection form takes even values. This is by the following string of arguments:
$M$ is spin iff $w_2 (TM)=0$.
$w_2 (TM)=0$ iff the linear form $H_2 (M; Z/2) \to Z/2$, $a \mapsto \langle w_2 (TM);a\rangle$ is null.
Any class $a \in H_2 (M;Z)$ can be represented as the fundamental class of an embedded oriented surface $F \subset M$.
$w_2 (TM)|_F = w_2 (\nu_F)$ by the product formula for Stiefel-Whitney classes and because $F$ is spin.
$w_2 (\nu_F)$ is the mod $2$ reduction of the Euler class of the normal bundle of $F$.
$\langle [F]; \chi(\nu_F) \rangle $ is the self-intersection number of $F$, or equivalently, the value of the quadratic form at $[F]$.
Now you should play a bit with $4$-manifolds and might get a feeling for the spin condition.
