New diode should "fix" it. Here's why. The diode is used to convert, or RECTIFY, the alternating current (AC) generated by the alternator, into direct current (DC) in order to keep the battery charged (because batteries CANNOT use AC). The alternators in most OPE applications have a very low amp output, and the max output is dependent upon engine RPM. The size of the alternator (as to AMP output) used in any specific engine is chosen by calculating the electrical load of the machine the engine will be used for. Matching the max amp output to the specific electrical load of the proposed machine means that a complex charging system (with high amp alternator AND a voltage regulator) are not needed. Fortunately for the consumer, all the calculating and choosing is done by the OEMs. In these "simple" charging systems, the diode that rectifies the AC into DC is selected with an "AMP Capacity" (ampacity) that is a bit greater than the max amp output of the alternator. An alternator with a max output of say, 3 AMP, might be provided with a charge diode that has an ampacity of 6 AMP. This means that the maximum amp output (3 AMP) of the alternator will NEVER overload the charge diode (6 AMP rated). The alternator charges a battery that has a potential amp output that can sometimes be nearly a "hundred X greater" than the "AMP rating" of the charge diode. This does not present a problem, nor cause an overload to the diode because a diode only allows current to flow in ONE DIRECTION (called "polarity bias", and in this case, the direction of current is FROM the alternator TO the BATTERY at a rate of 3 AMP. In this configuration, any current produced by the alternator, will be rectified by the diode and flow into the battery. When the engine stops, the alternator output stops too. At this point, the diode is just sitting there, inert, because there is no AC to rectify.
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Now, if you install the battery with reversed polarity, you have REVERSED the DIRECTION of current flow throughout the electrical wiring of the machine. In this configuration, the diode will attempt to flow current FROM the BATTERY to the ALTERNATOR. Because the copper wire windings of the alternator are relatively large, you can expect them to draw quite a lot of AMPS out of the battery. Remember, this particular charge diode is only rated at 6 AMP, but the amp draw from the battery now, (caused by connecting the battery directly to the alternator windings through the diode) might be 20 AMP, 40 AMP, 60 AMP, or higher, before the diode fails (or "blows"). The diode fails because the "bias" of the electrical system was changed to a configuration that exposed the diode to a much greater current than the measly 3 AMP of the alternator. That explains how the diode fails. What about the problem of the "engine will not stay running"? You tried to substitute a "solid conductor" (jumper wire) for a "biased conductor" (diode) in the charging circuit. In doing so, you are once again connecting the battery to the alternator windings, much the same as through the diode when the battery polarity was reversed. Except, now, there is no diode to "blow"! So the amp draw of the alternator is going to rise to the point that the main electrical system fuse will melt. When the fuse fails, the fuel solenoid on the carb is de-energized, so the engine stops.
When all things are connected and configured correctly, the alternator windings are a GENERATOR, or producer, of electricity. The diode effectively "separates" the alternator from the battery and only allows current to flow FROM the alternator, into the BATTERY. If you remove the diode and put a solid conductor in its place, the alternator windings become a CONSUMER of electricity, very much similar to a "resistance heater". It is the high resistance and subsequent "high amp draw" that causes the fuse to blow. If you replace both diode AND fuse, with a heavy piece of wire, we can then have a discussion about the nature of "FIRE".