If
$$\hat S_1=\frac{\hbar}{2}\pmatrix{0&1\\1&0}$$
is a spin matrix, i.e. a rank-$2$ tensor, and
$$\vec e_1=\pmatrix{1\\0\\0}$$
is the unit vector pointing in the $x$ direction, i.e. a rank-$1$ tensor,
then the tensor product $\hat S_1\otimes\vec e_1$ is a rank-$3$ tensor, since $2+1=3$. It may be abbreviated $\hat S_1\vec e_1$ and, yes, it may be written out as
$$\hat S_1\vec e_1 = \frac{\hbar}{2}\pmatrix{0&1\\1&0}\otimes\pmatrix{1\\0\\0} = \frac{\hbar}{2} \pmatrix{\pmatrix{0&1\\1&0}\\\pmatrix{0&0\\0&0}\\\pmatrix{0&0\\0&0}}.$$
It would be better to draw the tensor as a $2\times2\times3$ solid box of numbers, but that's hard to do on a chalkboard, so a matrix of matrices is the next best thing. If you search around the Internet, you can probably find better visualizations of rank-$3$ tensors.
Similarly, we have
$$\hat{\vec{S}}=\hat S_1\vec e_1+\hat S_2\vec e_2+\hat S_3\vec e_3
=\frac{\hbar}{2}\pmatrix{\pmatrix{0&1\\1&0}\\\pmatrix{0&0\\0&0}\\\pmatrix{0&0\\0&0}}
+\frac{\hbar}{2}\pmatrix{\pmatrix{0&0\\0&0}\\\pmatrix{0&-i\\i&0}\\\pmatrix{0&0\\0&0}}
+\frac{\hbar}{2}\pmatrix{\pmatrix{0&0\\0&0}\\\pmatrix{0&0\\0&0}\\\pmatrix{1&0\\0&-1}}
=\frac{\hbar}{2}\pmatrix{\pmatrix{0&1\\1&0}\\\pmatrix{0&-i\\i&0}\\\pmatrix{1&0\\0&-1}}.$$
You should not interpret $\hat S_1\vec e_1$ to mean matrix multiplication, in the sense of contracting one of the indices of $\hat S_1$ with the index of $\vec e_1$ to get a rank-$1$ tensor. (More bluntly, disregard Ivo Terek's suggestion.) Thankfully, since $\hat S_1$ has only two rows and two columns, there's no way to accidentally contract it with the three rows of $e_1$. As you point out in the comments, that doesn't even make sense.
In the bottom line, the middle dot "$\cdot$" is meant to denote a dot product, which gets rid of all the spatial indices by contracting them with each other, just like an ordinary dot product of two vectors in $3$-space. Specifically, the spatial index of $\hat{\vec{S}}$ is being contracted with the spatial index of $\vec e_r$. In symbols, $$\hat{\vec{S}}\cdot\vec e_r=\vec S_1\sin\theta\cos\phi+\vec S_2\sin\theta\sin\phi+\vec S_3\cos\theta.$$
This is a sum of three $2\times2$ spin matrices, which is again a $2\times2$ spin matrix.