I guess the answer in no when $\deg(X) \geq 3$.
In fact, in this case Griffiths showed that the two rulings $L_1$ and $L_2$ in $E$ are numerically equivalent in $\widetilde{X}$ (although not algebraically equivalent in general), so the corresponding $K_{\widetilde{X}}$-negative rays $R_1$ and $R_2$ in the Mori cone of $\widetilde{X}$ coincide.
It follows that the contraction of $R_1$ is actually the contraction of the whole quadric, that is it is not possible to contract the two rulings separately.
The situation is very subtle: in fact, the completed local ring $\widehat{\mathcal{O}}_{X,o}$ is not factorial, but the fact that $L_1$ is numerically equivalent to $L_2$ implies that $\mathcal{O}_{X,o}$ is factorial.
See [Debarre, Higher Dimensional Algebraic Geometry, p. 160] and the references given therein.
Note that without any assumption on the singularities the answer to your question is yes. For instance, consider the hypersurface $Y = \{x_0^2+x_1^3+x_2^4\}\subset\mathbb{P}^3$. Then $Sing(Y) = [0:0:0:1]$. The projective tangent cone of $Y$ in the singular point is not reduced. Therefore, the singularity is not ordinary. Let $\pi:X\rightarrow\mathbb{P}^3$ be the blow-up of $[0:0:0:1]$. It is easy to check that the singular locus of the strict transform $\widetilde{Y}\subset X$ contains a curve.
On the other hand, if the singularities are ordinary this can not happen.
More precisely:
Let $W\subset Z\subset X$ be smooth projective varieties, and let $Y\subset X$ be a projective variety such that $Sing(Y) = Z$ and $Y$ has ordinary singularities along $Z$. Let $\pi_W:X_W\rightarrow X$ be the blow-up of $W$, and $Z_W$, $Y_W$ the strict transforms of $Z$ and $Y$ respectively. Then $Sing(Y_W) = Z_W$ and $Y_W$ has ordinary singularities along $Z_W$.
Here is the proof:
The inverse image via $\pi_W$ of $Z$ is given by $Z_W\cup E_W$, where $E_W$ is the exceptional divisor of $\pi_W$. Now, we may consider the blow-up $\pi_{Z_w}:X_{Z_W}\rightarrow X_W$ of $Z_W\cup E_W$ in $X_W$, and the blow-up $\pi_Z:X_Z\rightarrow X$ of $Z$ in $X$. Note that since $E_W$ is a Cartier divisor in $X_W$ its blow-up does not produce any effect. By Corollary 7.15 in Harshorne there exists a morphism $f:X_{Z_W}\rightarrow X_Z$ such that $\pi_Z\circ f = \pi_W\circ \pi_{Z_{W}}$. Therefore, the morphism $f$ must map the exceptional divisor $E_{Z_W}$ of $\pi_{Z_W}$ onto the exceptional divisor $E_{Z}$ of $\pi_Z$. Furthermore, $f$ contracts the strict transform $\tilde{E}_W$ of $E_W$ in $X_{Z_W}$ onto $\pi_{Z}^{-1}(W)\subset E_Z$.
Now, let $g:\widetilde{X}_Z\rightarrow X_Z$ be the blow-up of $\pi_{Z}^{-1}(W)$. By the universal property of the blow-up, see Proposition 7.14 Hartshorne, there exits a unique morphism $\phi:X_{Z_W}\rightarrow \widetilde{X}_Z$ such that $g\circ\phi = f$.
Note that since we just blew-up smooth subvarieties both $X_{Z_W}$ and $\widetilde{X}_Z$ are smooth with Picard number $\rho(X_{Z_W}) = \rho(\widetilde{X}_Z) = \rho(X)+2$. In particular, the birational morphism $\phi:X_{Z_W}\rightarrow \widetilde{X}_Z$ can not be a divisorial contraction. Notice that $\phi$ can not be a small contraction either because otherwise $\widetilde{X}_Z$ would be singular. We conclude that $\phi$ is a bijective morphism, and since both $X_{Z_W}$ and $\widetilde{X}_Z$ are smooth $\phi$ is an isomorphism.
Now, let $Y_W$ be the strict transform of $Y$ in $X_W$, and assume that $Z_W\subsetneqq Sing(Y_W)$. Therefore, the strict transform $Y_{Z_W}$ of $Y_W$ in $X_{Z_W}$ is singular, and since $g$ is just the blow-up of a smooth variety the image $(g\circ \phi)(Y_{Z_W}) = f(Y_{Z_W})\subset X_Z$ is singular as well. Note that $f(Y_{Z_W})\subset X_Z$ is nothing but the strict transform of $Y$ via $\pi_Z$, and since $Sing(Y) = Z$ and $Y$ has ordinary singularities along $Z$ the blow-up $\pi_Z$ resolves the singularities of $Y$. A contradiction. We conclude that $Sing(Y_W) = Z_W$. Finally, if the intersection $Y_{Z_W}\cap E_{Z_W}$ is not transversal then the intersection $Y_Z\cap E_Z$ is not transversal as well. But this can not happen because the singularities of $Y$ are ordinary. Therefore, $Y_W$ has ordinary singularities along $Z_W$.
Best Answer
For the first question, and what Anton mentions, that the normal bundle of the exceptional divisor of the blow up of a smooth subvariety in a smooth variety is $\mathscr O(-1)$ see Thm 8.24 in [Hartshorne]. This takes care of your first question.
However, you have to be careful, because
To see that the normal bundle you are looking for (I assume, because what you wrote makes no sense) is the box product do this:
First let's give names to our players: Let $A$ denote the blow up of $\mathbb A^4$ at the origin, $P\simeq \mathbb P^3$ the exceptional divisor, $B\subset A$ the blow up of $\widetilde{\mathcal{X}}$ at the same point, and $E$ the exceptional divisor of that. Presumably you want $N_{E/B}$.
Observe that $A$ is smooth and hence all divisors are Cartier. Then consider the restrictions of $\mathscr O_A(P)$ to both $P$ and $B$ and then the restrictions of these to $E$:
$$ \mathscr O_A(P)|_P \simeq N_{P/A}\simeq \mathscr O_{\mathbb P^3}(-1) $$
$$ \left(\mathscr O_A(P)|_B\right)|_E \simeq \mathscr O_B(E)|_E \simeq N_{E/B} $$
and hence
$$ N_{E/B}\simeq \mathscr O_A(P)|_E \simeq \mathscr O_{\mathbb P^3}(-1)|_E\simeq \mathscr O_{\mathbb P^1}(-1)\boxtimes \mathscr O_{\mathbb P^1}(-1). $$