The chiral spinor representations of $\mathfrak{so}(2n)$ have dimension $2^{n-1}$. For $\mathfrak{so}(12)$ this is 32, not 64; $\mathfrak{so}(12)$ does not have an irrep with dimension 64.
Yamatsu has tables for $\mathfrak{so}(12)$ and $\mathfrak{so}(16)$. Note that the paper has 11232 pages, so it may take longer than you are expecting to download.
The tables in Slansky don't start until page 77. The preceding pages have a great deal of information about how to do calculations. Read the starting material in Yamatsu as well. If nothing makes sense, read a book on the representation theory of Lie algebras so that you understand Dynkin diagrams, Cartan matrices, roots and weights, Dynkin labels, the Weyl dimension formula, methods for decomposing tensor products, etc.
The basic flow for calculating dimensions is Dynkin diagram -> Cartan matrix -> simple roots -> root system -> positive roots -> Weyl dimension formula. Yamatsu gives an explicit but complicated form for the dimensions of $D_n$, based on the Weyl dimension formula.
Hand calculations for high-rank algebras are impractical; if you want to reproduce results by yourself you will probably need a computer algebra system to assist you. The Weyl dimension formula has many factors for the algebras you are asking about, because there are dozens of positive roots, so just determining the dimension for a given Dynkin label is tedious without a computer.
A simple, low-performance algorithm for computing tensor products is to compute the weight diagrams for the two irreps, add each weight in one to each weight in the other to get the weight diagram for their product, then decompose the weight diagram by finding the highest weight, subtracting all the weights associated with it, and repeating.
A better algorithm, used in the computer algebra implementations like LieART, is by Brauer and Klimyk. Humphreys' book describes it.
Slansky and Yamatsu describe simpler ways to decompose tensor products that can work if the algebra isn't too complicated.
I was able to reproduce Yamatsu's results for the $\mathfrak{so}(16)$ products you asked about by using these techniques:
The $\mathbf{128}$ has Dynkin label $(0,0,0,0,0,0,0,1)$ and the $\mathbf{128'}$ has label $(0,0,0,0,0,0,1,0)$. This means that the product $\mathbf{128}\otimes\mathbf{128}$ must contain the irrep with label $(0,0,0,0,0,0,0,2)$, which is the $\mathbf{6435'}$, and the product $\mathbf{128}\otimes\mathbf{128'}$ must contain the irrep with label $(0,0,0,0,0,0,1,1)$, which is the $\mathbf{11440}$.
Using Dynkin's second highest weight theorem, you can determine that $\mathbf{128}\otimes\mathbf{128}$ must also contain $(0,0,0,0,0,1,0,0)$, which is the $\mathbf{8008}$.
The irrep conjugacy numbers (see Yamatsu) are $(0,0)$ for $\mathbf{128}\otimes\mathbf{128}$ and $(0,2)$ for $\mathbf{128}\otimes\mathbf{128'}$. So the decomposition can contain only irreps with these conjugacy numbers.
Solve the Diophantine equations for the dimension and the Dynkin index. For $\mathbf{128}\otimes\mathbf{128}$ you need consider only the 4 irreps with conjugacy numbers $(0,0)$ and dimensions less than $128^2-6435-8008$. For $\mathbf{128}\otimes\mathbf{128'}$ you need consider only the 5 irreps with conjugacy numbers $(0,2)$ and dimensions less than $128^2-11440$.
It is possible that there is some simple and clever way to answer what you are asking. If so, I don't know it.
