tensor products in special realitivity Use the definition of a tensor as a linear function on vectors and 1-forms. For a one (2,1) tensor T, use a single vector as argument e.g. T(V) which converts it to a (1,1) tensor. However, if you write the original tensor as  a sum tensor product over tensor bases (two basis 1-forms and 1 basis vector) it is not clear which of the basis 1-forms acts on the argument --- there are two possibilities and the results are not the same unless T is  symmetric in the contravariant indices (not true in general). 
My question arises from the derivation in Schutz (First Course, First edition) on page 81, deriving Eq. 3.66 from Eq. 3.65.  He shows the derivative of a (1,1) tensor with respect to tau as the inner product with the velocity 4-vector and then claims that it is the contraction of a 1,2 tensor with that vector. It looks like the second basis vector (superscript gamma) is acting on U but it could also be acted on by the first (superscript beta). This would seem to be more consistent with the definition of the tensor product on page 71. Is there a convention here that I am missing?         
 A: Recall that the gradient of a scalar field is a (0,1) tensor, a one-form.
With a vector field $\vec A=\vec A(\vec r)$ you can think of having three scalar fields $P(\vec r),$ $Q(\vec r),$ and $R(\vec r)$ and then $\vec A(\vec r)=P(\vec r)\hat x+Q(\vec r)\hat y +R(\vec r)\hat z.$ We will do the same kind of thing in flat spacetime. Note that the vectors themselves do not change, only the scalar fields vary from place to place.
We know that the gradient of a scalar field is a one-form field. So we know how to take a directional derivative by contracting a unit vector with the gradient.
So now think of the (1,1) tensor field as being related to the 16 tensor fields $\tilde\omega^0\vec e_0,$ $\tilde\omega^0\vec e_1,$ $\tilde\omega^0\vec e_2,$ $\tilde\omega^0\vec e_3,$ $\tilde\omega^1\vec e_0,$ $\tilde\omega^1\vec e_1,$ $\tilde\omega^1\vec e_2,$ $\tilde\omega^1\vec e_3,$ $\tilde\omega^2\vec e_0,$ $\tilde\omega^2\vec e_1,$ $\tilde\omega^2\vec e_2,$ $\tilde\omega^2\vec e_3,$ $\tilde\omega^3\vec e_0,$ $\tilde\omega^3\vec e_1,$ $\tilde\omega^3\vec e_2,$ and $\tilde\omega^3\vec e_3.$ All we need is sixteen scalar fields $P,Q,R,S$ ... But that's a bit annoying so let's call the fields $T^\alpha_{\:\:\beta}$ so we can write $T=T^\alpha_{\:\:\beta}\tilde\omega^\beta\vec e_\alpha.$ But these are just sixteen ordinary scalar field multiplied by some (1,1) tensors that don't change from place to place.
So the difference between the two tensors at two places in spacetime is just given by knowing the difference between the 16 scalar fields at the two places and multiply each result by the (1,1) tensor $\tilde\omega^\beta\vec e_\alpha.$ And similarly to scale a tensor you can just scale the scalar field and multiply each result by the (1,1) tensor $\tilde\omega^\beta\vec e_\alpha.$ 
So far I've basically described how to use a  coordinate basis for a tensor.  But this tells us that
$$\frac{d T}{d\tau}=\left(\frac{d }{d\tau}T^\alpha_{\:\:\beta}\right)\tilde\omega^\beta\vec e_\alpha.$$
Where we are differentiating along a timelike curve parameterized by $\tau.$ But since $\frac{d}{d\tau}T^\alpha_{\:\:\beta}$ is just the directional derivative in a direction in spacetime we have that $\frac{d }{d\tau}T^\alpha_{\:\:\beta}=\frac{\partial T^\alpha_{\:\:\beta}}{\partial x^\gamma}\tilde\omega^\gamma (\vec U).$ Where $\vec U$ is the unit tangent to the worldline. This is just about how we take directional derivatives, they are equal to the unit tangents acting on the gradient.
OK. I deliberately out that partial as a fraction. So now we get to the question, which I think is just about notation.  If we denote $\frac{\partial T^\alpha_{\:\:\beta}}{\partial x^\gamma}$ by $T^\alpha_{\:\:\beta,\gamma}$ then we could write the basis tensor like $\vec e_\alpha\tilde\omega^\beta\tilde\omega^\gamma$ or we could write them like $\tilde\omega^\beta\tilde\omega^\gamma\vec e_\alpha$ and it doesn't matter because the order of a vector or a covector doesn't matter but we do care about the order of the two covectors and we write $\tilde\omega^\beta\tilde\omega^\gamma\vec e_\alpha$  with the $\beta$ first because we write $T^\alpha_{\:\:\beta,\gamma}$ with the $\beta$ first.
So we get $$\frac{d }{d\tau} T=\sum_{\alpha\beta\gamma}T^\alpha_{\:\:\beta,\gamma}\vec e_\alpha\tilde\omega^\beta\tilde\omega^\gamma\left(\sum_\delta U^\delta\vec e_\delta\right).$$
P.S. I have the second edition so I can't be sure I'm following you. But if you hit edit you can find out what I typed and use the same method to write your equations, then everyone can know what you are trying to say. The braces are grouping. The fraction has two arguments and the : is to put in a space, I out in two because the spaces looked too small to be clear.
