• The IC will try to turn the high-side switch on but it may fail because the bootstrap capacitor, C18 (used as a floating supply to drive the high-side switch at start up and throughout the normal operation), will have an initial voltage of 15 - 11.6 = 3.4 VDC (Ideally, the bootstrap capacitor could have charged up thanks to the empty therefore shorted output capacitor). If this is high enough then the high-switch will be turned on. If it is not then the IC may turn it off and turn the low-side switch on after some time as part of a stop-restart process (if there's any) and then turn it off again ⁽¹⁾. But there's still a possibility of that the circuit may not even start up.
• Following the turn-on of the high-side switch, the IC will fire up its internal soft-start which means increasing the internal reference voltage up to ~0.93 VDC gradually in 1 ms.

• Because the feedback voltage is already 0.76 VDC and the ref voltage is going to be way too low due to the soft-start, the IC will quite possibly stop switching (turn the high-side switch off) immediately as a response from the voltage loop. And then it'll turn the low-side switch on after a non-zero dead time.
• The IC will probably keep the low-side switch on until the end of the switching period because of the constant high voltage at the FB node, then turn it off. Turning the low-side switch off leads to the inductor's flyback action (note that the battery, the inductor, the low-side switch, and the body diode of the high-side switch form a boost regulator). So the SW node voltage will jump, and once it gets high enough the high-side switch body diode will clamp the SW node voltage to input voltage (~15 VDC). There could be a high clamping current through it, let alone the backdriven 15V source. Although this happens during the dead time only which is a few nanoseconds, there still must be a diode to block the flow of the current back to the 15V source. But let's assume nothing bad happens here.
• Whilst the high-side switch was on, the inductor voltage was V_L = 15 - 11.6 = 3.4 VDC (SW node +, output/battery node –) and the duration was quite possibly too low (presumably a few hundreds of nanoseconds, not much higher than the minimum on-time given in the datasheet). Whilst the low-side switch was on, the inductor voltage was 11.6 VDC and the duration would probably be much longer. This means a volt-second imbalance which leads to the failure of de-magnetisation of the core.
• Then the high-side switch will be turned on again. The internal ref voltage must have increased, but still less than 0.76 VDC. So the IC will stop switching again and the things above repeat.

The average current flowing through your 0.05 Ω shunt resistor will be negative (i.e. not to the battery, but from the battery) throughout the entire soft start period so it will never kick in and stop the process.

Not to mention the gradually-increasing current flow through the low-side switch which may kill the IC within a very short time because of the lack of low-side switch current limiting.

Your implementation may work for the applications that don't involve any battery charging. But still may lead to some strange behaviour.

It's better to use dedicated charger ICs that are readily available in the market. Those ICs have independent voltage and current loops, also they are optimised for longer life of the battery.

Your implementation may work for the applications that don't involve any battery charging. But still may lead to some strange behaviour.

It's best to use a dedicated charger IC. Many ICs are readily available in the market. They have independent voltage and current loops, also they are optimised for longer life of the battery.

• The IC will try to turn the high-side switch on but it may fail because the bootstrap capacitor, C18 (used as a floating supply to drive the high-side switch at start up and throughout the normal operation), will have an initial voltage of 15 - 11.6 = 3.4 VDC (Ideally, the bootstrap capacitor could have charged up thanks to the empty, large enough, and therefore shorted output capacitor). If this is high enough then the high-switch will be turned on. If it is not then the IC may turn it off and turn the low-side switch on after some time as part of a stop-restart process (if there's any) and then turn it off again ⁽¹⁾. But there's still a possibility of that the circuit may not even start up.
• Following the turn-on of the high-side switch, the IC will fire up its internal soft-start which means increasing the internal reference voltage up to ~0.93 VDC gradually in 1 ms.

• Because the feedback voltage is already 0.76 VDC and the ref voltage is going to be way too low due to the soft-start, the IC will quite possibly stop switching (turn the high-side switch off) immediately as a response from the voltage loop. And then it'll turn the low-side switch on after a non-zero dead time.
• The IC will probably keep the low-side switch on until the end of the switching period because of the constant high voltage at the FB node, then turn it off. Whilst the high-side switch was on, the inductor voltage was V_L = 15 - 11.6 = 3.4 VDC (SW node +, output/battery node –) and the duration was quite possibly too low (presumably a few hundreds of nanoseconds, not much higher than the minimum on-time given in the datasheet). Whilst the low-side switch was on, the inductor voltage was 11.6 VDC and the duration would be much longer. This means a volt-second imbalance which leads to the failure of de-magnetisation of the core.
• Turning the low-side switch off leads to the inductor's flyback action (note that the battery, the inductor, the low-side switch, and the body diode of the high-side switch form a boost regulator). So the SW node voltage will jump, and once it gets high enough the high-side switch body diode will clamp the SW node voltage to input voltage (~15 VDC). There could be a high clamping current through it, let alone the backdriven 15V source. Although this happens during the dead time only which is a few nanoseconds, there still must be a diode to block the flow of the current back to the 15V source. But let's assume nothing bad happens here.
• Then the high-side switch will be turned on again. The internal ref voltage must have increased, but still less than 0.76 VDC. So the IC will stop switching again and the things above repeat.

The average current flowing through your 0.05 Ω shunt resistor will be negative (i.e. not to the battery, but from the battery) and increasing in amplitude throughout the entire soft start period so your current limiter will never kick in and stop the process.

Not to mention the gradually-increasing amplitude of the current flowing through the low-side switch which may kill the IC within a very short time because of the lack of low-side switch current limiting.