I like Pot's argument better, but here's another approach for those who may be interested.
I get an upper bound of $2$, independent of $k$ and $m$.
Applying summation by parts, we have, for $k \geq 2$, $\begin{align*} \sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} &= \binom{m}{k} \sum_{i=1}^m \frac{1}{2^i} - \sum_{i=1}^{m-1} \left(\binom{i+1}{k} - \binom{i}{k}\right) \sum_{j=1}^i \frac{1}{2^j} \\ &= \binom{m}{k} \left(1 - \frac{1}{2^m}\right) - \sum_{i=1}^{m-1} \binom{i}{k-1} \left(1 - \frac{1}{2^i}\right)\\ &= \binom{m}{k} - \binom{m}{k}\frac{1}{2^m} - \sum_{i=1}^{m-1} \binom{i}{k-1} + \sum_{i=1}^{n-1} \binom{i}{k-1} \frac{1}{2^i}\\ &= \binom{m}{k} - \binom{m}{k}\frac{1}{2^m} - \binom{m}{k} + \sum_{i=1}^{m-1} \binom{i}{k-1} \frac{1}{2^i}\\ &= \sum_{i=1}^{m-1} \binom{i}{k-1} \frac{1}{2^i}- \binom{m}{k}\frac{1}{2^m}, \end{align*}$ where, in the second-to-last step, we use the upper summation identity for the binomial coefficients, $\displaystyle \sum_{i=0}^m \binom{i}{k} = \binom{m+1}{k+1}$.
Letting $\displaystyle F(m,k) = \sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i}$, this means we have the recurrence $F(m,k) = F(m-1,k-1) - \binom{m}{k}\frac{1}{2^m},$ valid for $k \geq 2$.
Unrolling the recurrence is easy, and with $F(m-k+1,1) = 2 - \frac{m-k}{2^{m-k+1}} - \frac{3}{2^{m-k+1}},$ we have
$\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} = 2 - \frac{m-k}{2^{m-k+1}} - \frac{3}{2^{m-k+1}} - \sum_{i=0}^{k-2} \binom{m-i}{k-i} \frac{1}{2^{m-i}}.$
Therefore, $\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} \leq 2.$