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Let $X$ be a closed, oriented smooth manifold of dimension $n$. Let $A,B$ be smooth submanifolds of dimensions $n-i$ and $n-j$ respectively. Assume that $A$ and $B$ intersect transversely, which means that for every $p\in A\cap B$, $$T_pA\oplus T_pB\to T_pX$$ is surjective.
Why does this map have to be surjective? For instance, if $A$ and $B$ are both straight lines in $\Bbb{R}^3$, and they intersect transversely, $T_pA\oplus T_pB\to T_p(\Bbb{R}^3)$ will not be surjective. The image will only be two dimensional.
The definition of transversality ensures that if $f\colon M \to Z$ and $g\colon N \to Z$ are smooth transverse functions, then $M\pitchfork N := f^{-1}(g(N))$ is a smooth manifold (i.e. the category of smooth manifolds admits pull-backs along transverse maps) and moreover if the intersection is non-empty then the co-dimension is preserved: $$dim(Z) - dim(N) = dim(M) - dim(M\pitchfork N)$$ In particular $dim(M\pitchfork N) = dim(M) + dim (N) - dim(Z)$ so if $M$ and $N$'s combined dimension is less than $Z$'s then "intersecting transversely" implies $f(M) \cap g(N) = \emptyset$. Normally you consider the case where $M$ and $N$ are submanifolds of $Z$, but $f$ and $g$ don't need to be embeddings.
There is a principle in differential topology that any two smooth maps $f\colon N\to Z$ and $g\colon M\to Z$ can be smoothly homotoped into a transverse pair of maps, so as above this means that if $dim(N) + dim(M) < dim(Z)$ then any pair $(f, g)$ can be smoothly homotoped into functions whose images don't intersect.
In particular you can use this to prove $\pi_k S^n \cong 0$ for $k < n$. Suppose that all spheres are based at $e_1$, and let $c\colon \{-e_1\}\to S^n$ denote the inclusion. Then given any pointed map $f\colon (S^k, e_1) \to (S^n, e_1)$, we can approximate it with a smooth map $f_s$ and then further by a map $f_t$ which is transverse to $c$, and therefore $-e_1$ is not in the image of $f_t$. Then there is a canonical null-homotopy of $f_t$ onto $e_1$.