For the Bernoulli numbers, take a look at Graham, Knuth, & Patashnik, Concrete Mathematics, Sections 6.5 and 7.6. In 6.5 they define the Bernoulli numbers $B_k$ by the implicit recurrence $\sum_{j=0}^m\binom{m+1}jB_j=[m=0]\tag{1}$ for all $m\ge 0$. (The righthand side is an Iverson bracket.) They then prove the identity $\sum_{k=0}^{n-1}k^m=\frac1{m+1}\sum_{k=0}^m\binom{m+1}kB_kn^{m+1-k}\;.$
In 7.6, on exponential generating functions, they rewrite $(1)$ by substituting $n$ for $m+1$ and adding $B_n$ to both sides to get $\sum_k\binom{n}kB_k=B_n+[n=1]\tag{2}$ for all $n\ge 0$. The lefthand side of $(2)$ is the binomial convolution of $\langle B_n:n\in\omega\rangle$ and the constant $1$ sequence. The egf of the constant $1$ sequence is just $e^z$; let $\widehat B(z)=\sum_{n\ge 0}B_n\frac{z^n}{n!}\;,$ the egf of $\langle B_n:n\in\omega\rangle$. Then $\sum_k\binom{n}kB_k=\widehat B(z)e^z\;.$ On the other hand, the egf of the righthand side of $(2)$ is $\sum_{n\ge 0}\Big(B_n+[n=1]\Big)\frac{z^n}{n!}=\widehat B(z)+z\;,$ so $\widehat B(z)e^z=\widehat B(z)+z\;,$ and $\widehat B(z)=\frac{z}{e^z-1}\;.$ The relationship with the Bernoulli polynomials is explored further in the next few pages.
I can’t help with the Euler (secant) numbers: I’ve only ever seen them defined as the coefficients in the Maclaurin expansion of $\operatorname{sech} x$. You might look at Exercise 7.41 and its solution, however, since it shows the connection between the up-down numbers and the tangent and secant functions.
Added: Suppose that $\widehat A(x)$ and $\widehat B(x)$ are exponential generating functions for $\langle a_n:n\in\omega\rangle$ and $\langle b_n:n\in\omega\rangle$, respectively, so that $\widehat A(x)=\sum_{n\ge 0}\frac{a_n}{n!}x^n\quad\text{ and }\quad\widehat B(x)=\sum_{n\ge 0}\frac{b_n}{n!}x^n\;.$
Now let $c_n=\sum_k\binom{n}ka_kb_{n-k}\;;$ the sequence $\langle c_n:n\in\omega\rangle$ is the binomial convolution of $\langle a_n:n\in\omega\rangle$ and $\langle b_n:n\in\omega\rangle$. Let $\widehat C(x)$ be the egf of this binomial convolution. Then
$\begin{align*} \widehat C(x)&=\sum_{n\ge 0}\frac{c_n}{n!}x^n\\ &=\sum_{n\ge 0}\sum_k\left(\frac{a_n}{k!}\cdot\frac{b_{n-k}}{(n-k)!}\right)x^n\\ &=\left(\sum_{n\ge 0}\frac{a_n}{n!}x^n\right)\left(\sum_{n\ge 0}\frac{b_n}{n!}x^n\right)\\ &=\widehat A(x)\widehat B(x)\;. \end{align*}$
Just as ordinary convolution of sequences is reflected in the product of their ordinary generating functions, binomial convolution of sequences is reflected in the product of their exponential generating functions.