There are two methods for determining the r-$\theta$ shear stress for your situation. Both methods give the same answer. One is to transform the equations for the stress tensor from Cartesian coordinates to cylindrical coordinates. This method is a little tedious for this problem.
The other method is to derive the equation for the stress tensor for your situation directly in cylindrical coordinates. The velocity vector is given here by: $$\vec{v}=v_{\theta}\vec{i}_{\theta}$$
The gradient (vector) operator is given by:
$$\vec{\nabla}=\vec{i}_r\frac{\partial}{\partial r}+\frac{\vec{i}_{\theta}}{r}\frac{\partial}{\partial \theta}$$
If we evaluate the gradient of the velocity vector, we obtain:
$$\vec{\nabla}\vec{v}=\vec{i}_r\frac{\partial (v_{\theta}\vec{i}_{\theta})}{\partial r}+\frac{\vec{i}_{\theta}}{r}\frac{\partial (v_{\theta}\vec{i}_{\theta})}{\partial \theta}=\vec{i}_r\vec{i}_{\theta}\frac{\partial v_{\theta}}{\partial r}-\vec{i}_{\theta}\vec{i}_r \frac{v_{\theta}}{r}$$
where we have assumed that $ v_{\theta}$ is not a function of $\theta$ and we have differentiated $\vec{i}_{\theta}$ with respect to $\theta$ to obtain $-\vec{i}_r$. The next step is to determine the rate of deformation tensor $\vec{E}$, which is the symmetric part of the velocity gradient tensor:
$$\vec{E}=(\vec{\nabla}\vec{v}+(\vec{\nabla}\vec{v})^T)/2=\frac{1}{2}\left[\frac{\partial v_{\theta}}{\partial r}- \frac{v_{\theta}}{r}\right](\vec{i}_r\vec{i}_{\theta}+\vec{i}_{\theta}\vec{i}_r )$$
The stress tensor $\vec{\tau}$ is equal to twice the viscosity $\eta$ times the rate of deformation tensor:
$$\vec{\tau}=\tau_{r\theta}\vec{i}_r\vec{i}_{\theta}+\tau_{\theta r}\vec{i}_{\theta}\vec{i}_r =\eta\left[\frac{\partial v_{\theta}}{\partial r}- \frac{v_{\theta}}{r}\right](\vec{i}_r\vec{i}_{\theta}+\vec{i}_{\theta}\vec{i}_r )$$
So, finally,
$$\tau_{r\theta}=\tau_{\theta r}=\eta\left[\frac{\partial v_{\theta}}{\partial r}- \frac{v_{\theta}}{r}\right]$$
Yikes. You're very confused.
Your first equation, which is incorrect, is trying to calculate the viscous stress vector (not shear stress vector) at the wall. In general, the viscous stress vector will have a component normal to the wall and a component tangent to the wall. The component tangent to the wall is called the shear stress. The correct equation for calculating the viscous stress vector exerted by the wall on the fluid is: $$\vec{\tau}=\mu \left[{\vec{\nabla} \vec{u}+(\vec{\nabla} \vec{u})^T}\right]\cdot \vec{n}$$where $\vec{n}$ is the unit normal vector drawn from the fluid through the wall. (This uses the sign convention that tensile stresses are positive and compressive stresses are negative)
The normal component of the viscous stress vector at the wall is obtained by dotting the stress vector $\vec{\tau}$ with the unit normal $\vec{n}$:$$\tau_n=\vec{\tau}\cdot\vec{n}$$
The shear component of the viscous stress vector at the wall is obtained by dotting the stress vector $\vec{\tau}$ with the unit tangent to the wall $\vec{t}$:$$\tau_t=\vec{\tau}\cdot\vec{t}$$So you can also write that:$$\vec{\tau}=\tau_n\vec{n}+\tau_t\vec{t}$$
The unit tangent vector is oriented parallel to the fluid velocity in close proximity to the wall. More generally, the shear stress vector at the wall, including direction, can be obtained by subtracting the normal component of the viscous stress vector from the total viscous stress vector:
shear stress vector $=\vec{\tau}-\tau_n\vec{n}=\vec{\tau}-(\vec{\tau}\cdot\vec{n})\vec{n} $
Hope this helps.
ADDENDUM
For the case of $\vec{u}=u_x\vec{i}_x+u_y\vec{i}_y+u_z\vec{i}_z$ and $\vec{n}=\vec{i}_y$, I get the following at the wall:
$$(\vec{\nabla}\vec{u})^T\cdot \vec{n}=\frac{\partial v_y}{\partial y}\vec{i_y}$$
$$(\vec{\nabla}\vec{u})\cdot \vec{n}=\frac{\partial v_x}{\partial y}\vec{i}_x+\frac{\partial v_z}{\partial y}\vec{i}_z+\frac{\partial v_y}{\partial y}\vec{i}_y$$
$$\vec{\tau}=\mu\left(\frac{\partial v_x}{\partial y}\vec{i}_x+\frac{\partial v_z}{\partial y}\vec{i}_z+2\frac{\partial v_y}{\partial y}\vec{i}_y\right)$$
I can see now what you're saying about the shear stress terms in $\vec{\tau}$, but there is also a normal stress component too. But you are correct that the shear stress component is parallel to the velocity vector near the wall.
Best Answer
Your assumptions are correct (but $r$ is often defined as the distance from the pipe centerline). However, this is a very specific case: laminar pipe flow.
In general, the stress will be a tensiorial quantity, defined as
$$ \tau_{ij}= \eta \frac{\partial u_i}{\partial x_j}$$
which is true for turbulent flow, in arbitrary geometries. Where $i,j$ are in the range ${1,2,3}$ for the $x,y,z$ components.
For your case, you only have velocities in the streamwise direction, and variations in the radial direction, which makes all other components zero.