In the classic Finite-Dimensional Vector Spaces by P. Halmos he defines the Tensor product as
The tensor product $U \otimes V$ of two finite-dimensional vector spaces $U$ and $V$ (over the same field) is the dual of the vector space of all bilinear forms on $U \oplus V$ [remark, by this he means all bilinear maps $\omega : U \times V \to \mathbb F$]. For each pair $x$ and $y$, with $x$ in $U$ and $y$ in $V$, the tensor product $z = x \otimes y$ of $x$ and $y$ is the element of $U \oplus V$ defined by $z(w) = w(x,y)$ for every bilinear form $w$.
Another construction, for example here or on Wikipedia, the tensor product is defined as the set of all formal sums on $U \times V$ modulo certain (''bilinear'') identities. This is equivalent to the set of all formal sums on $\{ u_i \}_{i\in I} \times \{ v_j \}_{j \in J}$ for bases $u_i$ and $v_j$ of $U$ and $V$.
According to my first reference, the set of all multilinear maps $V_1 \times \ldots \times V_k \to \mathbb F$ could be identified with $V_1^* \otimes \ldots \otimes V_k^*$, this seems always possible (not requiring finite dimensions), so in particular the set of all bilinear maps $U \times V \to \mathbb F$ could be identified with $V^* \otimes U^*$, so in Halmos definition the tensor product equals $(V^* \otimes U^*)^*$, where the tensor product is defined as the set of formals sums as above.
But to make sense, this definition must equal the definition by formal sums, so we must have $(V^* \otimes U^*)^* \cong V \otimes U$ to bring Halmos definition and the definition by formal sums in harmony.
But I guess here seems to be some sort of reflexivity involved, so he might used the finite-dimensionality of the vector spaces, or could Halmos definiton further extended, does I got it right?
Still another construction I found is by the following universal property (see also Wikipedia), i.e. the tensor product of two vector spaces $U$ and $V$ is such that for each bilinear map $U \times V \to W$ there exists exactly one linear map $U \otimes V \to W$, and this correspondence is linear, i.e. we have $$ \mbox{Bil}(U, V; W) \cong \mbox{Hom}(U\otimes V, W). $$ In our case $W = \mathbb F$ the ground field, then $$ \mbox{Bil}(U, V; \mathbb F) \cong (U\otimes V)^* $$ But by Halmos definition the tensor product equals $\mbox{Bil}(U, V; \mathbb F)^*$, so again this seems require some sort of reflexivity, i.e. $((U \otimes V)^*)^* \cong (U\otimes V)$.
But here the property to bring the definitions together reads still a little bit different, to make them all the same we need $(U^* \otimes V^*) \cong (U\otimes V)^*$, but this holds in general vector spaces, as $(U^* \otimes V^*) \cong \mbox{Bil}(U, V; F) \cong (U \otimes V)^*$.
So my questions again: 1) does I got everything right?, 2) am I right that Halmos definition is limited, and how far could we go with it, is reflexivity the right property to demand?
Yes, you've got it right. Remember the title of Halmos's book is Finite dimensional vector spaces! My guess is he probably thought the definition as the dual of bilinear forms was less abstract that either the quotient or the universal property definitions and used that one.
As you've realized Halmos's definition only agrees with the others when the spaces are reflexive, and since here we are talking about the algebraic dual (i.e., we don't have some norm and take dual to consist of bounded linear functionals, rather we are talking about all linear functionals), reflexive spaces are exactly the finite dimensional ones.