Show that the contracted multiplication U = ${{U_k}^j}_m$ = $S^{uj}T_{kum}$ is a (1,2)-tensor. Let $S^{ij}$ be the coordinates of a contravariant tensor S of the second order and $T_{klm}$ is the coordinates for a covariant tensor of third order. Show that the contracted multiplication U = ${{U_k}^j}_m$ = $S^{uj}T_{kum}$ is a (1,2)-tensor.
I want to show this by writing out the full tensor including the basisvectors.
I do understand that it is possible to show it by showing that the contracted tensor product transforms coordinates as a (1,2)-tensor, but I want to do it in a slightly other way, just by inspection of the expression of the tensor.
The coordinates S^{ij} are the product of two vectors and the coordinates $T_{klm}$ are the product of three covectors.
Contracting over the indices i and l , i = l = u we find :
$$ U = S^{uj}T_{kum}e_u \oplus e_j \oplus f^k \oplus f^u \oplus f^m $$
what I want my result to be is the following :
= ${{U_k}^j}_m f^k \oplus e_j \oplus f^m$
I do not understand the argument for how to get here. I understand that we sum over the indices u, but I dont understand why the two basisvectors $e_u$ and $f^u$ dissapear from the expression when we sum.
 A: You should be using $\otimes$, not $\oplus$, in the tensor product formulas.
Do you know how to think about vectors and covectors as being in a vector space $V$ and its dual space $V^*$? The tensor $S$ belongs to $V^{\otimes 2} = V \otimes V$ and the tensor $T$ belongs to $(V^*)^{\otimes 3} = V^* \otimes V^* \otimes V^*$. The idea of contraction is to just evaluate elements of $V^*$ on elements of $V$ in order to product scalars in a nice (bilinear) way.  The evaluation function $V \times V^* \to \mathbf R$ where $(v,\varphi) \mapsto \varphi(v)$ is a bilinear function of $v$ and $\varphi$ (linear in $v$ when $\varphi$ is fixed and linear in $\varphi$ when $v$ is fixed).  Therefore you get a linear function $V \otimes V^* \to \mathbf R$ that on elementary tensors $v \otimes \varphi$ has the value $\varphi(v)$.  Do you understand how bilinear or multilinear functions become linear functions when using tensor products?  This is the key point about tensor products of vector spaces: they linearize multilinear functions.
In any case, you are asking about how a contraction mapping
$$
V^{\otimes 2} \otimes (V^*)^{\otimes 3} \to V \otimes (V^*)^{\otimes 2}
$$
really works.  Specifically, you want to contract with the "first" piece of $V^{\otimes 2}$ and the "second" piece of $(V^*)^{\otimes 3}$ and you are confused about the basis vectors in those pieces disappear.  They do not really disappear, but rather you are evaluating the second piece of $(V^*)^{\otimes 3}$ on the first piece of $V^{\otimes 2}$ to get a scalar.  When you choose to use for your bases a basis of $V$ and its dual basis for the basis of $V^*$ then things appear to disappear because the very definition of a dual basis $e^1, \ldots, e^n$ in $V^*$ (you write $f^1, \ldots, f^n$) to a basis $e_1, \ldots, e_n$ of $V$ is the rule $e^i(e_j) = \delta_{ij}$, so $e^i(e_i) = 1$.
Here is how I would describe the contraction you ask about.  First we define the multilinear mapping
$$
V \times V \times V^* \times V^* \times V^* \to V  \otimes V^* \otimes V^*
$$
by combining the first coordinate from $V \times V$ and the second coordinate from $V^* \times V^* \times V^*$ using evaluation:
$$
(v_1,v_2,\varphi_1,\varphi_2,\varphi_3) \mapsto \varphi_2(v_1)v_2 \otimes \varphi_1 \otimes \varphi_3.
$$
The values of this function are multilinear in all 5 coordinates of $V \times V \times V^* \times V^* \times V^*$, so we get automatically a linear mapping
$$
V \otimes V \otimes V^* \otimes V^* \otimes V^* \to V  \otimes V^* \otimes V^*
$$
that has the following effect on elementary tensors:
$$
v_1 \otimes v_2 \otimes \varphi_1 \otimes \varphi_2 \otimes \varphi_3 \mapsto \varphi_2(v_1)v_2 \otimes \varphi_1 \otimes \varphi_3.
$$
This is what the contraction you write about really is.  Nothing is disappearing anywhere.  Suppose we pick a basis $e_1, \ldots, e_n$ of $V$ and use as a basis of $V^*$ the dual basis $e^1, \ldots, e^n$.  Then the 5-fold tensor product $V^{\otimes 2} \otimes (V^*)^{\otimes 3}$ has as a basis the 5-fold elementary tensors $e_i \otimes e_j \otimes e^k \otimes e^{\ell} \otimes e^m$.  The effect on these particular elementary tensors of your chosen contraction (evaluate 2nd dual tensorand in $(V^*)^{\otimes 3}$ on 1st tensorand in $V^{\otimes 2}$) is
$$
e_i \otimes e_j \otimes e^k \otimes e^{\ell} \otimes e^m \mapsto e^{\ell}(e_i) e_j \otimes e^k \otimes e^m = \delta_{i\ell} e_j \otimes e^k \otimes e^m.
$$
So if we have general tensors $S = \sum S^{ij} e_i \otimes e_j$ in $V^{\otimes 2}$ and $T = \sum T_{k\ell m} e^k \otimes e^\ell \otimes e^m$ in $(V^*)^{\otimes 3}$, then the above contraction $V^{\otimes 2} \otimes (V^*)^{\otimes 3} \to V \otimes (V^*)^{\otimes 2}$
has the following effect on $S \otimes T$:
\begin{eqnarray*}
S \otimes T = \sum_{ijk\ell m} S^{ij}T_{k\ell m} e_i \otimes e_j \otimes e^k \otimes e^\ell \otimes e^m  & \mapsto & \sum_{ijk\ell m} S^{ij}T_{k\ell m} \, e^\ell(e_i) \, e_j \otimes e^k \otimes e^\ell \otimes e^m  \\
& = & \sum_{ijk\ell m} S^{ij}T_{k\ell m}\delta_{i\ell} \, e_j \otimes e^k \otimes e^m  \\
& = & \sum_{ijk m} S^{ij}T_{k i m} e_j \otimes e^k \otimes e^m
\end{eqnarray*}
where the index $\ell$ goes away in the last calculation because if $\ell \not= i$ then the $(ijk\ell m)$-term is zero on account of $\delta_{i\ell} = 0$.
The coordinates of this final tensor are the numbers $S^{ij}T_{kim}$ in your post, but you chose to create a new index letter $u$ in place of $i$. Otherwise it is exactly the same thing.
When written in terms of a basis and dual basis, this contraction appears to be making things "disappear", but what is really going on is that you're just combining one tensorand (element of $V$) from $V^{\otimes 2}$ and one dual tensorand (in $V^*$) from $(V^*)^{\otimes 3}$, via evaluation, to produce a scalar so everything is depending multilinearly on the initial components.  If you did not use a dual basis in $V^*$ for calculations then the disappearing phenomenon would disappear: what's fundamental is evaluation of $V^*$ on $V$, a concept that does not depend on a choice of basis of the vector space or its dual space.
You say the tensor coordinates $S^{ij}$ are "products of two vectors" and $T_{k\ell m}$ are "products of three covectors". This is false: the coordinates are just numbers. They are not products of anything. The tensors $S$ and $T$ are in $V^{\otimes 2}$ and $(V^*)^{\otimes 3}$, but they are usually not elementary tensors $v \otimes v'$ or $\varphi \otimes \varphi' \otimes \varphi''$, so in no way are $S$ or $T$ a tensor product of two vectors or a tensor product of three covectors.
A simple analogy: polynomials in two variables are sums of monomials $c_{ij}x^iy^j$, but most polynomials in two variables are not products $g(x)h(y)$ of two single-variable polynomials.
Think about the dot product $\mathbf v \cdot \mathbf w$ on $\mathbf R^n$.  It is not making anything disappear: it's just a dot product.  But when you write it out in terms of the standard basis $\mathbf e_1, \ldots, \mathbf e_n$ then it looks like things disappear because $\mathbf e_i \cdot \mathbf e_j = \delta_{ij}$.
You are asking how to show the contracted tensor is a tensor. I did that by actually defining the contraction operation so it takes values in a tensor product space $V \otimes (V^*)^{\otimes 2}$, and thus the whole issue of "checking" the formula has values that are tensors goes away, just as it is tautological that the polyomial $1 - 2x + 7x^3$ is real-valued when $x$ is real.  What would you think if someone asked how to show the function $f : \mathbf R \to \mathbf R$ where $f(x) = 1 - 2x + 7x^3$ is real-valued? It's just automatic because the function is defined that way from the start.
