Example (property 1 fails, but property 2 is satisfied)
Look for $f$ as the blow up of an ideal sheaf $\mathscr I$, so $\widetilde X=\mathrm{Proj}_X(\oplus_d \mathscr I^d)$. Then the pre-image of the subscheme $Z\subset X$ defined by the ideal $\mathscr I$ is given by $\widetilde Z=\mathrm{Proj}_Y(\oplus_d \mathscr{I^d/I^{d+1}})$. Now if $X$ is Cohen-Macaulay and $Z$ is a complete intersection in $X$, (i.e., $\mathscr I$ is generated by a regular sequence), then $\mathscr{I/I^2}$ is locally free and
$\mathscr{I^d/I^{d+1}}\simeq \mathrm{Sym}^d(\mathscr{I/I^2})$ and hence $\widetilde Z\simeq \mathbb P(\mathscr{I/I^2})$.
Property #3 is kind of a red herring. The $(-1)$-twist is almost automatic, it comes from the construction of the blow up of $\mathscr I$.
Finally, here is a simple concrete example:
Let $X$ be a plane (or any smooth surface) and $\mathscr I=(x^2,y^2)$ where $x,y$ are local coordinates at a point. The blow up will be the surface with a pinch point (locally around the interesting singularity defined by $x^2z=y^2$) with the singular line contracted to a point. I think it is relatively easy to check that this satisfies properties #2 and #3.
To round things up Mike Roth in the comments below gives a nice example of a blow up along a non-smooth subvariety such that the resulting variety is actually smooth.
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
First of all, let $X$ be a smooth variety and $D$ an effective divisor on $X$. Denote $\operatorname{mult}_x(D)$ the multiplicity of $D$ at a point $x\in X$. The function $x\mapsto\operatorname{mult}_x(D)$ is known to be upper-semicontinuous on $X$. Therefore, if $Z\subset X$ is any irreducible subvariety, one can define the multiplicity of $D$ along $Z$, denoted $\operatorname{mult}_Z(D)$, to be the multiplicity $\operatorname{mult}_x(D)$ at a general point $x\in Z$.
Now, the second part of your question is just a matter of local computation. I will sketch it in the case where $X$ is a smooth surface and $Z$ is a point $x_0\in X$, leaving to you to work out the details in more general situations.
So, fix local coordinates $(x,y)$ centered at $x_0$ and consider the blow-up map $\mu\colon\widetilde X\to X$ given by $\mu(u,v)=(uv,v)$. In this chart, the exceptional divisor $E$ is given by the single equation $\{v=0\}$. Now, take a divisor $D\subset X$ whose local equation near $x_0$ is given by $\{f(x,y)=0\}$. Let $$ f(x,y)=\sum_{j,k\ge 0}a_{jk}x^jy^k. $$ Then, $\operatorname{mult}_{x_0}(D)=m$, where $m=\inf\{j+k\mid a_{jk}\ne 0\}$. Finally, a local equation for $\mu^*D$ is given by $$ f\circ\mu=f(uv,v)=\sum_{j,k\ge 0}a_{jk}u^jv^{j+k}=v^m\underbrace{\sum_{j,k\ge 0}a_{jk}u^jv^{j+k-m}}_{\text{holomorphic and does not vanish along $E$}}. $$ Thus, you see that $\mu^*D$ consist of the sum of one irreducible component given by $mE$ and the remaining part is just the proper transform of $D$.