Let me redo things with $f(x) = \frac{\delta^{-k}}{k!} \Delta_\delta^k g(x-k\delta)$, where $g(x) = (1-x)^k$ for $x\leq 1$ and $g(x)=0$ for $x>1$, and $\Delta h = h(y+\delta) - h(y)$. It is almost identical to the previous choice (with the multiplicative difference $\Delta_y^\cdot$) for $y = 1 + \delta$, $\delta$ small, and might be slightly more natural for the problem as stated. (Arguably, the version with $\Delta_y^\cdot$ is more natural for the problem that has $Mf(it)$ in place of $Mf(1+i t)$, not that that makes a big difference.) Some things become simpler and others more complicated.

I will give a variant at the end with $f(x) = \frac{\delta^{-k}}{k!} \Delta_\delta^k g(y-k\delta/2)$ (valid if we drop the condition that $f(x)=1$ for all $0\leq x\leq 1$); it gives better results.

Taking the Mellin transform of $f(x)$ is easy: $$\begin{aligned}Mf(s) &= \frac{\delta^{-k}}{k!} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} M(g(x - (k-j) \delta))(s) \\ &= \delta^{-k} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} \frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)}. \end{aligned}$$ The derivative of $\frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)}$ equals $$\frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)} \left(\log\left(1 + (k-j)\delta\right) - \frac{1}{s} - \frac{1}{s+1} - \dotsc - \frac{1}{s+k}\right)$$

Since $t\mapsto t^{k+1}$ is convex-up, $$\begin{aligned} \left|\sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} (1+ (k-j) \delta)^{s+k}\right|&\leq \sum_{j=0}^k \binom{k}{j} (1+(k-j) \delta)^{k+1}\leq 2^k \left(1 + \frac{k}{2} \delta\right)^{k+1} \end{aligned}$$ and similarly

In other words, $Mf(s) = \frac{C_{k}(s)}{s (s+1) \dotsc (s+k)}$, where $$|C_{k}(s)|\leq \left(\frac{2}{\delta}\right)^k \left(1 + \frac{k}{2} \delta\right)^{k+1} \sim \left(\frac{2}{\delta}\right)^k.$$ Hence

Let me redo things with $f(x) = \frac{\delta^{-k}}{k!} \Delta_\delta^k g(x-k\delta)$, where $g(x) = (1-x)^k$ for $x\leq 1$ and $g(x)=0$ for $x>1$, and $\Delta h = h(y+\delta) - h(y)$. It is almost identical to the previous choice (with the multiplicative difference $\Delta_y^\cdot$) for $y = 1 + \delta$, $\delta$ small, and the initial analysis is slightly easier. Let me also minimize the related quantity $$S_0 = |f(x)-1_{[0,1]}| + c \int_T^\infty |Mf(1+it)| dt$$ instead of the quantity of the statement: first, this is what actually needs to be minimized if one proves number-theoretical bounds in the traditional way (though I do need the quantity in the statement, personally); second, if we work with the quantity in the statement, there is a technical glitch that appears for this choice of $f$ and not for $f(x) = \frac{\delta^{-k}}{k!} (\Delta^\cdot_{1/y})^k \log^k(1/x)$. (If the solution to that glitch does not become apparent, then the latter choice of $f$ would indeed be preferrable.)

I will give a variant later with $f(x) = \frac{\delta^{-k}}{k!} \Delta_\delta^k g(y-k\delta/2)$ (valid if we drop the condition that $f(x)=1$ for all $0\leq x\leq 1$); it gives better results.

Taking the Mellin transform of $f(x)$ is easy: $$\begin{aligned}Mf(s) &= \frac{\delta^{-k}}{k!} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} M(g(x - (k-j) \delta))(s) \\ &= \delta^{-k} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} \frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)}. \end{aligned}$$ (Here comes the technical difficulty I referred to at first: the derivative of $\frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)}$ equals $$\frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)} \left(\log\left(1 + (k-j)\delta\right) - \frac{1}{s} - \frac{1}{s+1} - \dotsc - \frac{1}{s+k}\right).$$ The term $\log(1+(k-j) \delta)$ is undesirable; in fact, it makes the integral in the statement not converge for $k=1$.

Since, in general, $(Mf)'(s) = M((\log x) f(x))(s)$, it is not surprising that the "logarithmic" variant is more convenient as soon as we have $Mf'$ in the statement.)

Since $t\mapsto t^{k+1}$ is convex-up, $$\begin{aligned} \left|\sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} (1+ (k-j) \delta)^{s+k}\right|&\leq \sum_{j=0}^k \binom{k}{j} (1+(k-j) \delta)^{k+1}\\ &\leq 2^k \left(1 + \frac{k}{2} \delta\right)^{k+1}. \end{aligned}$$

In other words, $Mf(s) = \frac{C_{k}(s)}{s (s+1) \dotsc (s+k)}$, where $$|C_{k}(s)|\leq \left(\frac{2}{\delta}\right)^k \left(1 + \frac{k}{2} \delta\right)^{k+1} \sim \left(\frac{2}{\delta}\right)^k.$$ Hence $$\int_T^\infty |Mf(1+it)|dt \lesssim \left(\frac{2}{\delta}\right)^k \int_T^\infty \frac{dt}{t^{k+1}} = \frac{1}{k} \left(\frac{2}{\delta T}\right)^k.$$

Thus, we should choose $\delta$ so as to minimize $$S_0 = \frac{k \delta}{2} + \frac{c}{k} \left(\frac{2}{\delta T}\right)^k.$$ Taking derivatives, we see that the minimum is reached for $$\delta = \frac{2 (c/k)^{\frac{1}{k+1}}}{T^{\frac{k}{k+1}}};$$ then $$S_0 = (k+1) \left(\frac{c T}{k}\right)^{\frac{1}{k+1}} \frac{1}{T}.$$

Now we should choose $k$. Taking derivatives again, we see that we should have $1 + (k+1) \left(\frac{1}{k+1} \log \frac{c T}{k}\right)' = 1 - \frac{1}{k+1} \log \frac{c T}{k} - \frac{1}{k}$ close to $0$, i.e., $$k\approx \log \frac{c T}{k} + \frac{1}{k}.$$ Taking $k$ to be the integer closest to $$\log c T - \log \log c T$$ would do. Then we have $$\begin{aligned}S_0 &\approx (\log c T - \log \log c T + 1) \left(\frac{c T}{\log c T - \log \log c T}\right)^{\frac{1}{\log c T - \log \log c T}} \frac{1}{T} \\ &= \frac{e (\log c T - \log \log c T) + O\left(\frac{\log \log c T}{(\log c T)^2}\right)}{T}.\end{aligned}$$

Is this perhaps optimal up to a constant? Or up to its leading term?

(It does seem simple to show it is optimal up to a constant, and perhaps up to its leading term. To wit: we can consider a non-symmetric Büthe's "optimal" choice (2014), an integral of the Fourier transform of the function in Logan (1988), rescaled; I put "optimal" within quotation marks because Logan was really minimizing something related but different -- essentially, $\int_T^\infty |Mf(it)| dt$, with the constraint that $f(x)-1_{[0,1]}$ has support on a small interval $[e^{-\epsilon},e^{\epsilon}]$ -- but in the non-symmetric setting, the support does essentially give you $|f(x)-1_{[0,1]}|_1$, so Logan's function has to be essentially optimal in a class of function containing ours. If I say "perhaps" it is because all of this involves so many changes of variables that I still have to get right the constant in front of what one get from Logan!)