How do two tori glued along the identity of $S^1 \times S^1$ look like? I have got trouble imagining the outcome of gluing two solid tori $D^2 \times S^1 \sqcup_{\text{id}}S^1 \times D^2$ along the identity of $S^1 \times S^1$.
If I understand correctly, the tori are glued along their boundaries. How can one imagine that?
 A: There are actually many ways of doing this, depending on how we identify the boundaries of the two solid tori.  Any matrix in $SL_2(\mathbb{Z})$, defines a map $\mathbb{R}^2\to \mathbb{R}^2$, which is well defined modulo the lattice $\mathbb{Z}^2$.  That is, it induces a map $S^1\times S^1\to S^1\times S^1$.
Note the lattice points in $\mathbb{Z}^2$ are naturally identified with $\pi_1(S^1\times S^1)$ (as $\mathbb{Z}^2$ consists of the translates of the base point $(0,0)$ under the action of the fundamental group).  From your notation, it is natural to pick basis' $e_1,e_2$ for the fundamental group of the boundary of the first solid torus, and $f_1,f_2$ for the second, where $e_1$ and $f_2$ are each contractible in there respective solid tori.
It is easy to see that identifying boundaries via the map given by the matrix: $$\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$ results in $S^1\times S^2$.
The more interesting case is the identity matrix (which possibly is your original question):
$$\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$$
This gluing results in the three sphere $S^3$.  There are two natural ways to visualize this.
$$S^3=\{(x,y,z,w)\in \mathbb{R}^4\vert x^2+y^2+z^2+w^2=1\}$$
This decomposes into two solid tori: $$\left\{(x,y,z,w)\in S^3\vert x^2+y^2\leq \frac12\right\},\qquad \text{and} \qquad\left\{(x,y,z,w)\in S^3\vert z^2+w^2\leq \frac12\right\}.$$
The other way is to identify: $$S^3=\partial(D^2\times D^2)= S^1\times D^2 \sqcup_{s^1\times S^1} D^2\times S^1.$$
