My question is if this can be generalized as well: Let $\beta:\widetilde{X}\mathrel{\mathop:}=\mathop{\mathrm{Bl}}_Z(X)\to X$ be the blowup of a smooth algebraic variety $X$ along a irreducible generically smooth subscheme $Z$. Let $E\mathrel{\mathop:}=\beta^{1}(Z)$ be the exceptional divisor. Now, let us assume I have a divisor $D$ on $X$, with $Z$ closed subscheme of $D$ ($D$ smooth and irreducible, if it helps). Then I ask if $\beta^\ast D \sim \widetilde{D} + \alpha E$, for some integer $\alpha$, where $\widetilde{D}$ denotes the strict transform of $D$ and "$\sim$" is linear equivalence.
I think the answer is, almost. $E$ might have more than one component. Indeed, since you only assume that $Z$ is generically smooth, $E$ might have many different components. $E = \sum E_i$. You can always write $\beta^* D = \widetilde{D} + \sum_i \alpha_i E_i$, but you can't use the same $\alpha$ for all of them. Let me give an example. Suppose $X = \mathbb{A}^2$, $Z = V(x \cdot (x,y) \cdot (x,y^2)) = V(x^3, x^2y^2, x^2y, xy^3)$. Anyway, one chart of the blowup looks like $k[x/y, y^2/x]$ with the obvious map to $\text{Spec} k[x,y]$. On that chart, the two exceptional divisors are $E_1 = \text{Div}(x/y)$ and $E_2 =\text{Div}(y^2/x)$. On this chart, we compute $\beta^*D$. Note $$x^3 = ((x/y)^2 \cdot (y^2/x))^3$$ and so $$\beta^*D = \widetilde{D} + 6 E_1 + 3 E_2.$$ 

