From tensor seen as linear maps on vector spaces to index contraction I am considering a second-order tensor $\mathbf T$ seen as a linear maps on vector spaces, acting on a vector $\mathbf v$ to create a new vector $\mathbf u$, thus, $\mathbf u= \mathbf{T}(\mathbf{v})$. Starting from this very basic fact, how do we properly introduce the notation $\mathbf u= \mathbf{T}(\mathbf{v})=\mathbf{T}\cdot \mathbf{v}$ where the "$\cdot$" denotes the tensor product contracted once (or inner product, or scalar product when we deal with vectors)? In other words, how do we know that these two notations are interchangeable?
 A: If you are asking about notation, the comment of @anderstood is spot on. If $f:\mathbb{R}\rightarrow\mathbb{R}$ is any real function, writing $fx$ instead of $f(x)$ is confusing, because $fx$ could be the product of $f$ and $x$. However, if $f$ is a linear function (linear in a linear algebraic sense, eg. homogenous linear), then $f(x)=ax$, so identifying $f$ with $a$ gives us $f(x)=fx$.
The same notion is carried over into linear algebra, except in that case, $Tx$ is even less confusing, because there is no product operation on a general vector space and $T$ is NOT an element of the vector space anyways, so $Tx=T(x)$ is the only possible interpretation, even for nonlinear maps (unless you are explicitly identifying everything with matrices...).
So if we declare $T(x)\equiv T\cdot x$, this is a perfectly fine and unambigous thing to do.
If you are asking to prove if the action of $T$ on $x$ is the same as the contraction performed on $T\otimes x$ over the last two arguments, we can invoke the universal factorization property of tensor products, that essentially states that any multilinear function over the direct sum of some vector spaces has a unique representation as a linear map over the corresponding tensor spaces.
Now, if $T$ is seen as a linear operator on some finite dimensional vector space $V$ (over $\mathbb{F}$), then the space $\hom(V)$ of such maps is naturally isomorphic to the tensor product $V\otimes V^\ast$, and the action $T(x)$ can be seen as a bilinear map $(\cdot,\cdot):V\otimes V^\ast\times V\rightarrow V$. By the universal factorization property, there is a unique linear map, which we now call $\mathrm{tr}^2_1:V\otimes V^\ast\otimes V\rightarrow V$ so that $(T,x)=T(x)=\mathrm{tr}^2_1(T\otimes x)$, which is what we call "contraction over the first lower index and second upper index".
By this argument, any tensor action can be seen as a contraction performed over a tensor product.
A: Given a $(1,1)$ tensor as a map
$$
\tau\colon V\times V^*\to\mathbb{C}
$$
such element can be always expressed as $\tau = \tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\otimes \alpha^{\nu}$, with $(e_{\mu}, \alpha^{\nu})$ being a basis of $V, V^*$, respectively.
Using your notation, let us act with $\tau$ upon a vector $v\in V$. We have
$$
\tau(v) = \left(\tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\otimes \alpha^{\nu}\right)(v)= \left(\tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\right)\cdot\alpha^{\nu}(v) = \left(\tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\right)\alpha^{\nu}(v^{\lambda}e_{\lambda})=\tau^{\mu}_{\phantom{\mu}\nu}v^{\lambda}\delta^{\nu}_{\phantom{\nu}\lambda}e_{\mu}
$$ 
which reduces to $\tau^{\mu}_{\phantom{\mu}\lambda}v^{\lambda}e_{\mu}\in V$. The action of the $(1,1)$ tensor $\tau$ on $v$ is denoted, with abuse of notation, as $\tau(v)=(v,\tau)=\tau\cdot v$.
Have a look at this other answers of mine here and here as well.
