First, $\mathbf{F} = x\mathbf{\hat i} + y\mathbf{\hat j} + z\mathbf{\hat k}$ converted to spherical coordinates is just $\mathbf{F} = \rho \boldsymbol{\hat\rho} $. This is because $\mathbf{F}$ is a radially outward-pointing vector field, and so points in the direction of $\boldsymbol{\hat\rho}$, and the vector associated with $(x,y,z)$ has magnitude $|\mathbf{F}(x,y,z)| = \sqrt{x^2+y^2+z^2} = \rho$, the distance from the origin to $(x,y,z)$.
You also asked about where
$$\begin{bmatrix}\boldsymbol{\hat\rho} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\phi} \end{bmatrix} = \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\phi & \cos\phi & 0 \end{bmatrix} \begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}$$
comes from. Let's look at the simpler 2D case first. For a point $(x,y)$ it helps to imagine that you're on a circle centered at the origin. In this case, the two fundamental directions you can move are perpendicular to the circle or along the circle. For the perpendicular direction we use the outward-pointing radial unit vector $\mathbf{\hat{r}}$. For the other direction, moving along the circle means (instantaneously) that you're moving tangent to it, and we take the unit vector in this case to be $\boldsymbol{\hat\theta}$, pointing counterclockwise. For example, suppose you're at the point $(1/\sqrt{2},1/\sqrt{2})$. Then, in the graph below, $\mathbf{\hat{r}}$ is in red and $\boldsymbol{\hat\theta}$ is in yellow.
Note that this means that, unlike the unit vectors in Cartesian coordinates, $\mathbf{\hat{r}}$ and $\boldsymbol{\hat{\theta}}$ aren't constant; they change depending on the value of $(x,y)$.
Now, what about a formula for $\mathbf{\hat{r}}$? If we move perpendicular to the circle we're keeping $\theta$ fixed in the polar coordinate representation $(r \cos \theta, r \sin \theta)$. The vector $\mathbf{\hat{r}}$ is the unit vector in the direction of this motion. If we interpret $r$ as time, taking the derivative with respect to $r$ will give us the velocity vector, which we know points in the direction of motion. Thus we want the unit vector in the direction of $\frac{d}{dr} (r \cos \theta, r \sin \theta) = (\cos \theta, \sin \theta)$. This is already a unit vector, so $\mathbf{\hat{r}} = \cos \theta \mathbf{\hat{x}} + \sin \theta \mathbf{\hat{y}} $. Similarly, moving counterclockwise along the circle entails keeping $r$ fixed in the polar coordinate representation $(r \cos \theta, r \sin \theta)$. Thus to find $\boldsymbol{\hat\theta}$ we take $\frac{d}{d\theta} (r \cos \theta, r \sin \theta) = (-r \sin \theta, r \cos \theta)$. This is not necessarily a unit vector, and so we need to normalize it. Doing so yields $\boldsymbol{\hat\theta}= -\sin \theta \mathbf{\hat{x}} + \cos \theta \mathbf{\hat{y}} $. In matrix form, this is $\begin{bmatrix} \mathbf{\hat{r}} \\ \boldsymbol{\hat\theta}\end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \mathbf{\hat{x}} \\ \mathbf{\hat{y}} \end{bmatrix}$.
Moving up to spherical coordinates, for a given point $(x,y,z)$, imagine that you're on the surface of a sphere. The three fundamental directions are perpendicular to the sphere, along a line of longitude, or along a line of latitude. The first corresponds to $\boldsymbol{\hat\rho}$, the second to $\boldsymbol{\hat\theta}$, and the third to $\boldsymbol{\hat{\phi}}$. (This is using the convention in the Wikipedia page, which has $\theta$ and $\phi$ reversed from what you have.) Thus to find $\boldsymbol{\hat\rho}$, $\boldsymbol{\hat\theta}$, and $\boldsymbol{\hat{\phi}}$, we take the derivative of the spherical coordinate representation $(\rho \sin \theta \cos \phi, \rho \sin \theta \sin \phi, \rho \cos \theta)$ with respect to $\rho$, $\theta$, and $\phi$, respectively, and then normalize each one. That's where the matrix
$$\begin{bmatrix}\boldsymbol{\hat\rho} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\phi} \end{bmatrix} = \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\phi & \cos\phi & 0 \end{bmatrix} \begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}$$
comes from.
As Henning Makholm points out, one way to view what we're doing here is that we're rotating the $\mathbf{\hat{x}}, \mathbf{\hat{y}}, \mathbf{\hat{z}}$ vectors. The transformation matrix can thus be considered a change-of-basis matrix. This means you could also (and more generally) convert $\mathbf{F} = x\mathbf{\hat i} + y\mathbf{\hat j} + z\mathbf{\hat k}$ to spherical coordinates via
\begin{align}
\mathbf{F}
&=
\begin{bmatrix} F_{\rho}\\ F_{\theta}\\ F_{\phi}\end{bmatrix}
\\
&=
\begin{bmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -
\sin \theta \\ - \sin \phi & \cos \phi & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}
\\
&=
\begin{bmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & - \sin \theta \\ - \sin \phi & \cos \phi & 0 \end{bmatrix} \begin{bmatrix} \rho \sin \theta \cos \phi \\ \rho \sin \theta \sin \phi \\ \rho \cos \theta \end{bmatrix}
\\
&=
\begin{bmatrix} \rho \sin^2 \theta \cos^2 \phi + \rho \sin^2 \theta \sin^2 \phi + \rho \cos^2 \theta \\ \rho \sin \theta \cos \theta \cos^2 \phi + \rho \sin \theta \cos \theta \sin^2 \phi - \rho \sin \theta \cos \theta \\ - \rho \sin \theta \sin \phi \cos \phi + \rho \sin \theta \sin \phi \cos \phi + 0 \end{bmatrix}
\\
&= \begin{bmatrix} \rho \\ 0 \\ 0\end{bmatrix}.
\end{align}
So we get $\mathbf{F} = \rho \boldsymbol{\hat\rho} + 0 \boldsymbol{\hat\theta} + 0 \boldsymbol{\hat{\phi}} = \rho \boldsymbol{\hat{\rho}}$, just as we argued at the beginning.
A far more simple method would be to use the gradient.
Lets say we want to get the unit vector $\boldsymbol { \hat e_x } $. What we then do is to take $\boldsymbol { grad(x) } $ or $\boldsymbol { ∇x } $.
This; $\boldsymbol ∇ $, is the nabla-operator. It is a vector containing each partial derivative like this...
$\boldsymbol { ∇= ( \frac {∂} {∂x}, \frac {∂} {∂y}, \frac {∂} {∂z}) } $
When we take the gradient of x we get this...
$\boldsymbol { ∇x= ( \frac {∂x} {∂x}, \frac {∂x} {∂y}, \frac {∂x} {∂z})=(1,0,0)=\hat e_x } $
To get the unit vector of $\boldsymbol x$ in cylindrical coordinate system we have to rewrite $x$ in the form of $\boldsymbol {r_c}$ and $\boldsymbol {\phi}$.
$\boldsymbol {x= r_c cos(x) } $
Now you have to use the more general definition of nabla ($\boldsymbol ∇ $).
Lets say we have a curve-linear coordinate system where the position vector is defined like this...
$\boldsymbol {\vec r = u_1 \hat e_{u1} + u_2 \hat e_{u2} + u_3 \hat e_{u3}} $
... Then the nabla operator for that coordinate system is as follows...
$\boldsymbol { ∇ = \frac {1}{h_1} \frac {∂}{∂u_1} \hat e_{u1} + \frac {1}{h_2} \frac {∂}{∂u_2} \hat e_{u1} + \frac {1}{h_3} \frac {∂}{∂u_3} \hat e_{u1}} $
"$\boldsymbol { h_n } $" is the scale factor to the variable "$\boldsymbol { u_n } $". The scale-factor is defined as: $\boldsymbol {h_n = \frac {\partial \vec r}{\partial u_n}}$
For cylindrical coordinates the position vector is defined as: $\boldsymbol {\vec r = r_c \hat e_{rc} + z \hat e_z }$
With some simple math we can get the scale factors and they are...
$\boldsymbol {h_{rc} = 1 \ \ ,\ h_{\phi} = r_c \ \ ,\ h_z = 1}$
We already know that in cylindircal cooridnates $\boldsymbol x $ is defined as $\boldsymbol {x = r_c cos(x)}$, so now we can get the gradient.
$\boldsymbol { ∇x = ∇(r_c cos(x))= \frac {\partial (r_c \cos(x))}{\partial r_c} \hat e_{rc} + \frac {1}{r_c} \frac {\partial (r_c \cos(x))}{\partial \phi} \hat e_{\phi} + \frac {\partial (r_c cos(x))}{\partial z} \hat e_{z}}$
The result from this gradient is then...
$\boldsymbol {\hat e_{x} = \cos(\phi)\hat e_{rc} - \sin(\phi) \hat e_{\phi}}$
When the same method is applied to $\boldsymbol y$, where $\boldsymbol {y = r_c sin(\phi) }$, we get with ease that...
$\boldsymbol {\hat e_{y} = sin(\phi)\hat e_{rc} + \cos(\phi) \hat e_{\phi}}$
Hope it helped!
I also hope the use of $\boldsymbol \phi $ instead of $\boldsymbol \theta $ and $\boldsymbol {r_c} $ instead of $\boldsymbol \rho $ wasn't to confusing. As a physics student I am more used to the $\boldsymbol {(r_c,\phi,z)}$ standard for cylindrical coordinates.
Best Answer
Since $\phi$ is constant the conversion is straightforward, indeed we simply have that