You almost got it. The amount of numbers between $1$ and $100$ that are divisible by some $n$ such that $n \in \mathbb{Z}_{+}$, is $\lfloor \frac{100}{n} \rfloor$ ($\lfloor x \rfloor$ denotes the floor function, which round any number down to the nearest integer, or as you say it, removes the decimals). Therefore the probability of choosing a number, divisible by $n$, between $1$ and $100$ is $\frac{\lfloor \frac{100}{n} \rfloor}{100}$. Therefore your steps should have followed as so
Event $\mathcal{F_1}$: Choosing a number from $1$ and $100$ divisible by $5$. $P(\mathcal{F_1})= \frac{\lfloor \frac{100}{5} \rfloor}{100} = \frac{\lfloor 20 \rfloor}{100} = \frac{20}{100} = \frac{1}{5}.$
Event $\mathcal{F_2}$: Choosing a number from $1$ and $100$ divisible by $7$. $P(\mathcal{F_2}) = \frac{\lfloor \frac{100}{7} \rfloor}{100} = \frac{\lfloor 14 \rfloor}{100} = \frac{14}{100} = \frac{7}{50}.$
It seems the answer is solving for the exclusive or (notation: $\oplus$), therefore we must account for intersection, $35$ and $70$, probability $P(\mathcal{F_1} \cap \mathcal{F_2}) = \frac{2}{100} = \frac{1}{50}.$
Therefore, $P(\mathcal{F_1 \oplus F_2}) = P(\mathcal{F_1}) + P(\mathcal{F_2}) - P(\mathcal{F_1} \cap \mathcal{F_2}) = \frac{1}{5} + \frac{7}{50} - \frac{1}{50} = \frac{8}{25}.$ Or simply notice that $P(\mathcal{F_1 \oplus F_2}) = \frac{\frac{\lfloor \frac{100}{5} \rfloor}{100} + \frac{\lfloor \frac{100}{7} \rfloor}{100} - \frac{2}{100}}{100} = \frac{20 + 14 - 2}{100} = \frac{8}{25}.$
Make sure that you at least check your cases. For example, when you say the probability for numbers divisible $7$ is $\frac{7}{100}$, you're claiming that are are $7$ numbers that are divisible by $7$ between $1$ and $100$. This can easily be disproven by just checking $8$ numbers. $7, 14, 21, 28, 35, 42, 49, \text{and }56$ are all divisible by $7$ and less then $100$.
