There is a parameterized curve $X(\lambda)$. It is expanded at $\lambda=\lambda_0+\epsilon$
$$X(\lambda)=(X\vert_{\lambda_0}+\epsilon\frac{dX}{d\lambda}\vert_{\lambda_0}+\frac{1}{2}\epsilon^2\frac{d^2X}{d\lambda^2}\vert_{\lambda_0}+\cdots)$$ X is taken out of the parentheses leaving the "operator" inside $$X(\lambda)=(1+\epsilon\frac{d}{d\lambda}+\frac{1}{2}\epsilon^2\frac{d^2}{d\lambda^2}+\cdots)X\vert_{\lambda_0}$$
$$(1+\epsilon\frac{d}{d\lambda}+\frac{1}{2}(\epsilon\frac{d}{d\lambda})^2+\cdots)$$ is said to be equal to Taylor expansion of $exp(\epsilon \frac{d}{d\lambda})$ that is known
$$exp(\epsilon \frac{d}{d\lambda})=(1+\epsilon\frac{d}{d\lambda}+\frac{1}{2}\epsilon^2\frac{d^2}{d\lambda^2}+\cdots)$$ So $X(\lambda)=exp(\epsilon \frac{d}{d\lambda})X\vert_{\lambda_0}$
It means $\frac{d^2}{d\lambda^2}=(\frac{d}{d\lambda})^2$. The square of differentiation is equal to second-order differentiation. How is this should be understood?
Treat $\frac{d} {d\lambda} $ as an abstract operator, $T$ say.
Define repeated application of $T$ on a function $X$ as follows: $(T)^2 X = T (T X) $. In other words T of T of X.
Write T in terms of derivatives and you see the result you wanted