blow-ups of subvarieties $Y_1,Y_2$ are disjoint in the blow-up at the ideal $I(Y_1)+I(Y_2)$ Exercise 9.19 in Gathmann's notes reads:

Let $X\subset \Bbb A^n$ be an affine variety, and let $Y_1,Y_2$ be irreducible closed subsets of $X$ , none contained in the other. We blow-up X along the ideal $\Bbb I(Y_1)+\Bbb I(Y_2)$ to get $\tilde X$. Show that the strict transforms of $Y_1$ and of $Y_2$ are disjoint in $\tilde X$.

Let's say $\Bbb I(Y_1)=(f_1,...,f_r)$, $\Bbb I(Y_2)=(g_1,...,g_s)$. Then $\Bbb I(Y_1)+\Bbb I(Y_2)=(f_1,...,f_r,g_1,...,g_s)$ and we define a morphism $f: X \to \Bbb P^{r+s-1}$ by $x \mapsto (f_1(x),...,g_s(x))$. Let $\Gamma_f$ be its graph, then $\tilde X$ is the closure of $\Gamma_{f}$ in $X \times \Bbb P^{r+s-1}$, and 
we identify the blow-up of $Y_1$ at $\Bbb I(Y_1)+\Bbb I(Y_2)$ (i.e. the strict transform of $Y_1$) with the closure of $\Gamma_{f|_{Y_1}}$ in 
$X \times \Bbb P^{r+s-1}$ and similarly we do for $Y_2$. Then it seems strange to claim that these two closures are disjoint in $\tilde X$, since $\Gamma_{f|_{Y_1}}$, $\Gamma_{f|_{Y_2}}$ themselves do not seem to be disjoint. Indeed, if $a \in Y_1\cap Y_2$ then $(a,f(a)) \in \Gamma_{f|_{Y_1}} \cap \Gamma_{f|_{Y_2}}$, allegedly contradicting the claim. 
Where is my mistake? Also, what is the geometric meaning of this claim?
 A: Keeping your notation, the blow up can be described as follows. Let $A$ be the coordinate ring of $X$, i.e. $X=\mathrm{Spec}A$, and $I=I(Y_1)+I(Y_2)=I(Y_1 \cap Y_2)$. Consider the map of rings
\begin{equation}
\varphi: A[y_1,\ldots,y_r,z_1,\ldots,z_s] \rightarrow \bigoplus_{d \geq 0} I^d,
\end{equation}
where $I^0=A$, and $\varphi(y_i)=f_i$, $\varphi(z_j)=g_j$. The direct sum $\bigoplus_{d\geq0}I^d$ is a ring. It is a quotient of $A[y_1,\ldots,y_r,z_1,\ldots,z_s]$, since there are algebraic relations between the $f_i$'s and $g_j$'s, while the polynomial variables $y_i$'s and $z_j$'s have no interesting relations. If you are not familiar with this concept, try with the blow up of $\mathbb{A}^2$ at the origin.
Now, by definition $\tilde{X}=\mathrm{Proj}(\bigoplus_{d\geq0}I^d)$, and the surjective map of rings realizes it as a closed subscheme of $\mathbb{P}^{r+s-1}_X$. In particular, $ker(\varphi)$ gives you the ideal in $A[y_1,\ldots,y_r,z_1,\ldots,z_s]$ defining the blow up as subscheme.
Now, the strict transforms of $Y_1$ and $Y_2$ are nothing but the blow up of $Y_1$ and $Y_2$ at suitable ideals: $(I_1+I_2)/I_1 \subset A/I_1$ and $(I_1+I_2)/I_2 \subset A/I_2$, respectively. Clearly, $(I_1+I_2)/I_1$ is generated by $(\bar{g}_1,\ldots,\bar{g_s})$, since quotienting by $I_1$ kills the $f_i$'s. Analogously, $(I_1+I_2)/I_2$ is generated by the images of the $f_i$'s.
Now, we want to describe the two missing blow ups as before. Thus, we get
\begin{equation}
\psi_1:A/I_1[z_1,\ldots,z_s] \rightarrow \bigoplus_{d\geq 0} ((I_1+I_2)/I_1)^d,
\end{equation}
where $((I_1+I_2)/I_1)^0=A/I_1$, and
\begin{equation}
\psi_2:A/I_2[y_1,\ldots,z_r] \rightarrow \bigoplus_{d\geq 0} ((I_1+I_2)/I_2)^d,
\end{equation}
where $((I_1+I_2)/I_2)^0=A/I_2$.
Notice that $\mathbb{P}^{s-1}_{Y_1}$ and $\mathbb{P}^{r-1}_{Y_2}$ are naturally subschemes of $\mathbb{P}^{r+s-1}_{X}$, defined by $(f_1, \ldots, f_s,y_1,\ldots,y_r)$ and $(g_1, \ldots, g_s,z_1,\ldots,z_s)$. Notice that their intersection is given by the ideal $(f_1, \ldots, f_s,g_1, \ldots, g_r, y_1,\ldots,y_r,z_1,\ldots,z_s)$. This ideal contains the irrelevant ideal $(y_1,\ldots,y_r,z_1,\ldots,z_s)$, therefore it corresponds to the empty set. So, inside $\mathbb{P}^{r+s-1}_X$, we have $\mathbb{P}^{s-1}_{Y_1} \cap \mathbb{P}^{r-1}_{Y_2} = \emptyset$. Since $\tilde{Y}_1 \subset \mathbb{P}^{s-1}_{Y_1}$ and $\tilde{Y}_2 \subset \mathbb{P}^{r-1}_{Y_2}$, we get $\tilde{Y}_1 \cap \tilde{Y}_2 = \emptyset$.
Now, what does it mean from the geometric point of view? The ideal $I_1+I_2$ corresponds to the intersection $Y_1 \cap Y_2$. What does blowing up a locus $Z$ mean? It roughly means that we replace $Z$ by a divisor. Over the smooth locus of $Z$ it exactly correspond to considering a projective bundle of rank $\mathrm{codim}Z-1$ (this for instance tells you that nothing happens if we blow up a smooth divisor). Now, roughly speaking, what happens is that we replace $Y_1 \cap Y_2$, where $Y_1$ and $Y_2$ are supposed to meet, with something of bigger dimension. This makes so that "$Y_1$ and $Y_2$ have room to miss each other". This is made precise by the fact that the projective bundle that dominates the intersection $Y_1 \cap Y_2$ parametrizes the tangent directions to the intersection itself. What I have just said needs interpretation: in case $Y_1$ and $Y_2$ meet properly, then what happens is literally what I wrote. In case they are tangent, then the blow up will get some singularity or reduced structure that will take care of separating the two strict transforms (since the tangent directions might agree in this case).
As advice, try to work out the two following examples. First, consider the two lines $y=x$ and $y=-x$ in $\mathbb{A}^2$, and see what happens if you blow up their intersection. This illustrates the nice case. Then, try to do the same with $y=x^2$ and $y=-x^2$. This corresponds to the nastier case.
Notice one thing: above we used that the strict transform of $Y_1$ under the blow up of $Y_1 \cap Y_2$ is the same as the blow up of $Y_1 \cap Y_2$ in $Y_1$. This works because $Y_1 \cap Y_2$ is a subscheme of $Y_1$. In general the picture is more complicated. For instance, I told you that if we blow up a smooth divisor we are doing nothing. Thus, if we blow up $y=0$ in $\mathbb{A}^2$ we get back $\mathbb{A}^2$. In particular, the strict transform of the nodal curve $y^2=x^2(x+1)$ is the nodal curve itself. On the other hand, the intersection of the two is the nodal point, and if you blow up the nodal curve at the singular point you get a smooth curve.
A: Note that the statement is true without requiring that neither subvariety is contained in the other. Write $I_j = I(Y_j)$ for $j = 1,2$. Note that $V(I_1+I_2) = Y_1\cap Y_2$. If $Y_1\subset Y_2$, then $Y_1\subset V(I_1+I_2) $, i.e. $Y_1\setminus V(I_1+I_2) = \varnothing$. Using Gathmann's definition, clearly the fact that $Y_1\setminus V(I_1+I_2) = \varnothing$ implies that the strict transform of $Y_1$ in $\widetilde X$ is empty. But then it is also (trivially) disjoint from $Y_2$.
I think Gathmann includes the condition because he doesn't want the reader to think of this trivial situation.
Here is an elementary proof (using Gathmann's definitions) that the strict transforms $\widetilde Y_1$ and $\widetilde Y_2$ in $\widetilde X = \text{blow-up of }X\, \text{at }I_1+I_2$ are
disjoint:
Proof. Write $I_1 + I_2 = (f_1,\dots ,f_{r+s})$ such that $I_1 = (f_1,\dots ,f_r)$ and $I_2 = (f_{r+1},\dots ,f_{r+s})$. Let $U = X\setminus V(I_1+I_2)$. For each $(x,\lbrack y\rbrack)\in Y\cap U$ (square brackets = homogenous coordinates) we then have $\lbrack y_1,\dots ,y_{r+s}\rbrack = \lbrack f_1(x),\dots ,f_{r+s}(x)\rbrack$, so there is nonzero $\lambda$ with $\lambda y_i = f_i(x)$ for $i=1,\dots ,r+s$. But since $x\in Y_1 = V(I_1)$, we have $f_i(x) = 0$ for $i = 1,\dots ,r$, hence $y_i = 0$ for $i =1,\dots ,r$. This polynomial equation also holds in the closure $\widetilde Y_1 = \overline{Y_1\cap U}$. Similarly $(x,\lbrack y\rbrack)\in\widetilde Y_2$ must have $y_i = 0$ for $i = r+1,\dots ,r+s$. But then $(x,\lbrack y\rbrack)\in \widetilde Y_1\cap \widetilde Y_2$ must have all homogenous coordinates $y_1 = \dots = y_{r+s} = 0$, which is impossible. Q.E.D.
