I think I've heard this statement before but I'd like to make sure it's true.
Let $X$ be a variety and $L$ a line bundle on it. Take $S < P\left(H^0(X,L)\right)$ to be a linear subspace of the projective space of sections. Such a thing is called a linear system and produces a (rational) map $X \to P\left( H^0(X,L)^\vee \right)$ to the dual projective space, sending $x$ to $\epsilon_x \colon H^0(X,L) \to k$ which takes $s$ to $\epsilon_x(s) = s(x)$.
The baselocus $A$ of the linear system $S$ is given by $\{x \in X | s(x)=0, \forall s \in S\}$. The universal family is $H$, given by $\{(s,x) \in S \times X | s(x) = 0 \}$.
My question is: is $H = Bl_A X$? If so, why?
The answer to your question is no. Consider the case $X = \mathbb P^n$ mapping to itself by the identity (so $S = P(H^0(X,\mathcal O(1)))$). Then $H \subset S \times X$ is the set of pairs of a point $x$ and a hyperplane $s$ s.t. $s(x) = 0$ (and $A$ is the empty set).
An easy dimension count gives the dimension of $H$: for each $x$, there is a $\mathbb P^{n-1}$ of hyperplanes passing through $x$, so dim $H = 2n - 1$. So unless $n = 1$, $X$ and $H$ aren't of the same dimension, and so $H$ can't be a blow-up of $X$.
But in any case, I think you might have been trying to formulate a different statement:
The map $X \to P(S^{\vee})$ is not defined at $A$, so we get a birational map from $X$ to the Zariski closure of the image of $X \setminus A$. Sometimes, this lifts to an isomorphism between $Bl_A X$ and this Zariski closure.