Per request, here is a more detailed answer:
We first note that, for any bounded operator $T$, the kernel of $T^*$ is the orthogonal complement of the range of $T$. Indeed, $x \in \mathrm{ker}(T^*)$ iff $T^*(x) = 0$, iff $\langle T^*(x), y \rangle = 0, \forall y$, iff $\langle x, T(y) \rangle = 0, \forall y$, iff $x \perp \mathrm{range}(T)$. Now, $T_k^*T_j = 0$ implies $\mathrm{range}(T_j) \subseteq \mathrm{ker}(T_k^*) = \mathrm{range}(T_k)^\perp$, so the ranges of $T_k$ are orthogonal to each other.
Let $H_k = \mathrm{ker}(T_k)^\perp = \overline{\mathrm{range}(T_k^*)}$. Since $T_kT_j^* = 0$, we have $\mathrm{range}(T_j^*) \subseteq \mathrm{ker}(T_k) = \mathrm{range}(T_k^*)^\perp$. Taking closures, we see that $H_k = \overline{\mathrm{range}(T_k^*)}$ are orthogonal to each other. Hence, we may write $H$ as an orthogonal direct sum $H = (\oplus_k H_k) \oplus H'$, where $H'$ is orthogonal to all $H_k$. For any $f \in H$, we may then write $f = \sum_k f_k + f'$ where $f_k \in H_k$, $f' \in H'$. Observe that, for any fixed $k$, for $j \neq k$, $f_j \in H_j \subseteq H_k^\perp = (\mathrm{ker}(T_k)^\perp)^\perp = \mathrm{ker}(T_k)$, so $T_k(f_j) = 0$. Similarly, $T_k(f') = 0$. This implies that $T_k(f) = T_k(f_k)$.
Now, we have $S_N(f) = \sum_{k = -N}^N T_k(f) = \sum_{k = -N}^N T_k(f_k)$. As we have observed, the ranges of $T_k$ are orthogonal to each other, so $||S_N(f)||^2 = \sum_{k = -N}^N ||T_k(f_k)||^2$. Since $||T_k|| \leq 1$ for all $k$, we have $||S_N(f)||^2 \leq \sum_{k = -N}^N ||f_k||^2$. Note that terms in $f = \sum_k f_k + f'$ are orthogonal to each other, so $||f||^2 = \sum_k ||f_k||^2 + ||f'||^2 \geq \sum_{k = -N}^N ||f_k||^2$. Thus, $||S_N(f)||^2 \leq ||f||^2$. As this holds for all $f$, $||S_N|| \leq 1$. (This will be useful later in showing $||T|| \leq 1$.)
We now show that $S_N(f)$ is Cauchy. Indeed, assume $N > M$, then,
$$||S_N(f) - S_M(f)||^2 = \sum_{k = -N}^{-M - 1} ||T_k(f_k)||^2 + \sum_{k = M + 1}^N ||T_k(f_k)||^2 \leq \sum_{k = -\infty}^{-M - 1} ||T_k(f_k)||^2 + \sum_{k = M + 1}^{\infty} ||T_k(f_k)||^2 \leq \sum_{k = -\infty}^{-M - 1} ||f_k||^2 + \sum_{k = M + 1}^{\infty} ||f_k||^2$$
Observe that $\sum_k ||f_k||^2 \leq ||f||^2 < \infty$. Thus, for any $\epsilon > 0$, for large enough $M$, $\sum_{k = -\infty}^{-M - 1} ||f_k||^2 + \sum_{k = M + 1}^{\infty} ||f_k||^2 < \epsilon$, whence $||S_N(f) - S_M(f)||^2 < \epsilon$. Thus, $S_N(f)$ is a Cauchy sequence, so it converges to some element of $H$, which we shall denote by $T(f)$. Since $S_N$ are all linear, properties of limits easily imply that $T$ is a linear operator. Since $||S_N|| \leq 1$, $||S_N(f)|| \leq ||f||$, so $||T(f)|| = \lim_{N \rightarrow \infty} ||S_N(f)|| \leq ||f||$, that is, $||T|| \leq 1$.
The following is an example which you might find illuminating. It also demonstrates that $S_N$ might not converge in the norm topology. Consider $H = l^2(\mathbb{Z})$. Let $T_k$ be the rank one orthogonal projection onto the space spanned by the standard basis vector $e_k$. Then $S_N$ is the orthogonal projection onto $\mathrm{span}(e_{-N}, \cdots e_N)$. From this you can check that $||S_N - S_M|| = 1$ whenever $N \neq M$, so $S_N$ does not converge in the norm topology. But for any fixed $f$, $S_N(f)$ does converge. In fact, the sequence converges to $f$ itself, so $T$, as we constructed, is simply the identity operator.
