Solution:

By the binomial theorem,
$$(a + 101)^{n} \equiv a^{n} + \binom{n}{1} a^{n - 1}101 = a^{n} + 101n a^{n - 1} \pmod {101^{2}}.$$
Using this gives (all congruences are modulo $101^{2}$)
$$\begin{align*} f(a + 101) &= 100!\sum_{n = 0}^{100}\frac{(a + 101)^{n}}{n!} \\ &\equiv 100!\sum_{n = 0}^{100}\left(\frac{a^{n}}{n!} +\frac{101n a^{n - 1}}{n!}\right) \\ &\equiv f(a) + 100!\cdot 101\sum_{n = 1}^{100}\frac{a^{n - 1}}{(n - 1)!} \\ &\equiv f(a) + 101f(a) - 100!\cdot 101\frac{a^{100}}{100!} \\ &\equiv f(a) + 101(f(a) - 1) \\ &\equiv 20 + 101(20 - 1) = \boxed{1939}\quad (\mathrm{mod}\ 101^{2}). \end{align*}$$

Solution:

The above solution can be viewed as a consequence of Hensel's lemma as follows. Because $101$ is prime, for any integer $x$ not divisible by $101$, we have that
$$f^{\prime}(x) = 100!\left(1 + x + \frac{x^{2}}{2!} +\dots +\frac{x^{99}}{99!}\right) = f(x) - x^{100}\equiv f(x) - 1\pmod {101}.$$
Clearly $101 \nmid a$. Hence, by Hensel's lemma, we get that
$$f(a + 101) \equiv f(a) + 101f^{\prime}(a) \equiv 20 + 101 \cdot 19 \equiv \boxed{1939} \pmod {101^{2}}.$$